Projectiles

2 Calculate the range on a horizontal plane of a small stone projected from a point on the plane with speed \(12 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of elevation of \(27 ^ { \circ }\).

1 A particle is projected with speed \(15 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of \(40 ^ { \circ }\) above the horizontal from a point on horizontal ground. Calculate the time taken for the particle to hit the ground.

1. Given that \(\tan \theta = \frac{4}{3}\), find the horizontal distance travelled by \(P\) when it first reaches three-quarters of its greatest height. Give your answer in terms of \(u\) and \(g\). [4]

6 A particle \(P\) is projected from a point \(O\) on horizontal ground. 0.4 s after the instant of projection, \(P\) is 5 m above the ground and a horizontal distance of 12 m from \(O\).

1. Calculate the initial speed and the angle of projection of \(P\).
2. Find the direction of motion of the particle 0.4 s after the instant of projection.

6 A particle \(P\) is projected from a point \(O\) on horizontal ground. 0.4 s after the instant of projection, \(P\) is 5 m above the ground and a horizontal distance of 12 m from \(O\).

1. Calculate the initial speed and the angle of projection of \(P\).
2. Find the direction of motion of the particle 0.4 s after the instant of projection.

5.(a)The box below shows a student's attempt to prove the following identity for \(a > b > 0\) $$\arctan a - \arctan b \equiv \arctan \frac { a - b } { 1 + a b }$$ Let \(x = \arctan a\) and \(y = \arctan b\) ,so that \(a = \tan x\) and \(b = \tan y\) $$\begin{aligned} \text { So } \tan ( \arctan a - \arctan b ) & \equiv \tan ( x - y ) \\ & \equiv \frac { \tan x - \tan y } { 1 - \tan ^ { 2 } ( x y ) } \\ & \equiv \frac { a - b } { 1 - ( a b ) ^ { 2 } } \\ & \equiv \frac { a - a b + a b - b } { ( 1 - a b ) ( 1 + a b ) } \\ & \equiv \frac { a ( 1 - a b ) - b ( 1 - a b ) } { ( 1 - a b ) ( 1 + a b ) } \\ & \equiv \frac { a - b } { 1 + a b } \end{aligned}$$ Taking arctan of both sides gives \(\arctan a - \arctan b \equiv \arctan \frac { a - b } { 1 + a b }\) as required. There are three errors in the proof where the working does not follow from the previous line.

1. Describe these three errors.
2. Write out a correct proof of the identity.
   (b)[In this question take \(g\) to be \(9.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }\) ] \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{4d5b914c-28b2-4485-a42e-627c95fa16e2-22_244_1267_1870_504} \captionsetup{labelformat=empty} \caption{Figure 3}
   \end{figure} Balls are projected,one after another,from a point,\(A\) ,one metre above horizontal ground. Each ball travels in a vertical plane towards a 6 metre high vertical wall of negligible thickness,which is a horizontal distance of \(10 \sqrt { 2 }\) metres from \(A\) . The balls are modelled as particles and it is assumed that there is no air resistance.
   Each ball is projected with an initial speed of \(28 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and at a random angle \(\theta\) to the horizontal,where \(0 < \theta < 90 ^ { \circ }\) Given that a ball will pass over the wall precisely when \(\alpha \leqslant \theta \leqslant \beta\)
3. find, in degrees, the angle \(\beta - \alpha\)
4. Deduce that the probability that a particular ball will pass over the wall is \(\frac { 2 } { 3 }\)
5. Hence find the probability that exactly 2 of the first 10 balls projected pass over the wall. You should give your answer in the form \(\frac { P } { Q ^ { k } }\) where \(P , Q\) and \(k\) are integers and \(P\) is not a multiple of \(Q\).
6. Explain whether taking air resistance into account would increase or decrease the probability in (b)(iii).
7. find, in degrees, the angle \(\beta - \alpha\)

A stone is projected from a point \(O\) on horizontal ground. The equation of the trajectory of the stone is $$y = 1.2x - 0.15x^2,$$ where \(x\) m and \(y\) m are respectively the horizontal and vertically upwards displacements of the stone from \(O\). Find

1. the greatest height of the stone, [2]
2. the distance from \(O\) of the point where the stone strikes the ground. [2]

1 A small ball is projected with speed \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of \(45 ^ { \circ }\) above the horizontal from a point \(O\) on horizontal ground. At time \(t \mathrm {~s}\) after projection, the horizontal and vertically upwards displacements of the ball from \(O\) are \(x \mathrm {~m}\) and \(y \mathrm {~m}\) respectively.

1. Express \(x\) and \(y\) in terms of \(t\).
2. Show that the equation of the trajectory of the ball is \(y = x - \frac { 1 } { 40 } x ^ { 2 }\).
3. State the distance from \(O\) of the point at which the ball first strikes the ground.

14 A particle \(P\) is projected from the point \(O\), at the top of a vertical wall of height \(H\) above a horizontal plane, with initial speed \(V\) at an angle \(\alpha\) above the horizontal. At time \(t\) the coordinates of the particle are \(( x , y )\) referred to horizontal and vertical axes at \(O\).

1. Express \(x\) and \(y\) as functions of \(t\). Let \(\theta\) be the angle \(O P\) makes with the horizontal at time \(t\).
2. (a) Show that $$\tan \theta = \tan \alpha - \frac { g } { 2 V \cos \alpha } t$$ (b) Show that when the particle attains its greatest height above the point of projection, where \(O P\) makes an angle \(\beta\) with the horizontal, $$\tan \beta = \frac { 1 } { 2 } \tan \alpha .$$ (c) If the particle strikes the ground where \(O P\) makes an angle \(\beta\) below the horizontal, show that $$H = \frac { 3 V ^ { 2 } \sin ^ { 2 } \alpha } { 2 g }$$

4 A small ball \(B\) is projected from a point 1.5 m above horizontal ground with initial speed \(29 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of \(30 ^ { \circ }\) above the horizontal.

1. Show that \(B\) strikes the ground 3 s after projection.
2. Find the speed and direction of motion of \(B\) immediately before it strikes the ground.

7 An arrow is fired from a point at a height of 1.5 metres above horizontal ground. It has an initial velocity of \(12 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of \(30 ^ { \circ }\) above the horizontal. The arrow hits a target at a height of 1 metre above horizontal ground. The path of the arrow is shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{f30b02da-a41e-44cb-b45f-9e6a3a9d0528-18_341_1260_550_390} Model the arrow as a particle.

1. Show that the time taken for the arrow to travel to the target is 1.30 seconds, correct to three significant figures.
2. Find the horizontal distance between the point where the arrow is fired and the target.
3. Find the speed of the arrow when it hits the target.
4. Find the angle between the velocity of the arrow and the horizontal when the arrow hits the target.
5. State one assumption that you have made about the forces acting on the arrow.
   (1 mark)
   \includegraphics[max width=\textwidth, alt={}]{f30b02da-a41e-44cb-b45f-9e6a3a9d0528-19_2486_1714_221_153}
   \includegraphics[max width=\textwidth, alt={}]{f30b02da-a41e-44cb-b45f-9e6a3a9d0528-20_2486_1714_221_153}

\includegraphics{figure_3} A rocket \(R\) of mass 100 kg is projected from a point \(A\) with speed 80 m s\(^{-1}\) at an angle of elevation of \(60°\), as shown in Fig. 3. The point \(A\) is 20 m vertically above a point \(O\) which is on horizontal ground. The rocket \(R\) moves freely under gravity. At \(B\) the velocity of \(R\) is horizontal. By modelling \(R\) as a particle, find

1. the height in m of \(B\) above the ground, [4]

1. the time taken for \(R\) to reach \(B\) from \(A\). [2]

When \(R\) is at \(B\), there is an internal explosion and \(R\) breaks into two parts \(P\) and \(Q\) of masses 60 kg and 40 kg respectively. Immediately after the explosion the velocity of \(P\) is 80 m s\(^{-1}\) horizontally away from \(A\). After the explosion the paths of \(P\) and \(Q\) remain in the plane \(OAB\). Part \(Q\) strikes the ground at \(C\). By modelling \(P\) and \(Q\) as particles,

1. show that the speed of \(Q\) immediately after the explosion is 20 m s\(^{-1}\), [3]

1. find the distance \(OC\). [6]

END

1 A particle is projected horizontally with speed \(12 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from the top of a high cliff. Find the direction of motion of the particle after 2 s .

1 A particle is projected horizontally with speed \(12 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from the top of a high cliff. Find the direction of motion of the particle after 2 s .

2 One end of a light inextensible string of length 0.8 m is attached to a fixed point, \(O\). The other end is attached to a particle \(P\) of mass \(1.2 \mathrm {~kg} . P\) hangs in equilibrium at a distance of 1.5 m above a horizontal plane. The point on the plane directly below \(O\) is \(F\). \(P\) is projected horizontally with speed \(3.5 \mathrm {~ms} ^ { - 1 }\). The string breaks when \(O P\) makes an angle of \(\frac { 1 } { 3 } \pi\) radians with the downwards vertical through \(O\) (see diagram). \includegraphics[max width=\textwidth, alt={}, center]{0428f2f2-12c4-4e89-93ab-8cfe2c5aca4a-02_757_889_1482_251}

1. Find the magnitude of the tension in the string at the instant before the string breaks.
2. Find the distance between \(F\) and the point where \(P\) first hits the plane.

7 In this question the \(x\) - and \(y\)-directions are horizontal and vertically upwards respectively and the origin is on horizontal ground.
A ball is thrown from a point 5 m above the origin with an initial velocity \(\binom { 14 } { 7 } \mathrm {~ms} ^ { - 1 }\).

1. Find the position vector of the ball at time \(t \mathrm {~s}\) after it is thrown.
2. Find the distance between the origin and the point at which the ball lands on the ground.

4 A projectile P travels in a vertical plane over level ground. Its position vector \(\mathbf { r }\) at time \(t\) seconds after projection is modelled by $$\mathbf { r } = \binom { x } { y } = \binom { 0 } { 5 } + \binom { 30 } { 40 } t - \binom { 0 } { 5 } t ^ { 2 } ,$$ where distances are in metres and the origin is a point on the level ground.

1. Write down
   (A) the height from which P is projected,
   (B) the value of \(g\) in this model.
2. Find the displacement of P from \(t = 3\) to \(t = 5\).
3. Show that the equation of the trajectory is $$y = 5 + \frac { 4 } { 3 } x - \frac { x ^ { 2 } } { 180 } .$$

8. [In this question, the unit vectors \(\mathbf { i }\) and \(\mathbf { j }\) are in a vertical plane, with \(\mathbf { i }\) being horizontal and \(\mathbf { j }\) being vertically upwards.] \begin{figure}[h] \end{figure} At time \(t = 0\), a small ball is projected from a fixed point \(O\) on horizontal ground. The ball is projected from \(O\) with velocity ( \(p \mathbf { i } + q \mathbf { j }\) ) \(\mathrm { ms } ^ { - 1 }\), where \(p\) and \(q\) are positive constants. The ball moves freely under gravity. At time \(t = 3\) seconds, the ball passes through the point \(A\) with velocity ( \(8 \mathbf { i } - 12 \mathbf { j }\) ) \(\mathrm { m } \mathrm { s } ^ { - 1 }\), as shown in Figure 4.

1. Find the speed of the ball at the instant it is projected from \(O\). For an interval of \(T\) seconds the speed, \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), of the ball is such that \(v \leqslant 10\)
2. Find the value of \(T\). At the point \(B\) on the path of the ball, the direction of motion of the ball is perpendicular to the direction of motion of the ball at \(A\).
3. Find the vertical height of \(B\) above \(A\).

4 A particle is projected with velocity \(V\), at an angle of elevation of \(60 ^ { \circ }\) to the horizontal, from a point on a plane inclined at an angle of \(30 ^ { \circ }\) to the horizontal. The path of the particle is in a vertical plane through a line of greatest slope. If \(R _ { 1 }\) and \(R _ { 2 }\) are the respective ranges when the particle is projected up the plane and down the plane, show that $$R _ { 2 } = 2 R _ { 1 }$$

4 A plane is inclined at an angle \(\theta ^ { \circ }\) to the horizontal. A particle is projected from a point A on the plane with speed \(V \mathrm {~ms} ^ { - 1 }\) in a direction making an angle of \(\phi ^ { \circ }\) with a line of greatest slope of the plane. The particle lands at a point B on the plane, as shown in the diagram, and the time of flight is \(T\) seconds. \includegraphics[max width=\textwidth, alt={}, center]{feb9a438-26b0-41d3-b044-6acd6efccde0-4_332_872_461_246} \begin{enumerate}[label=(\alph*)] \item By considering the motion of the particle perpendicular to the plane, show that \(\mathrm { T } = \frac { 2 \mathrm {~V} \sin \phi } { \mathrm {~g} \cos \theta }\). Consider the case when \(\theta = 30 , \phi = 25\) and \(V = 20\). \item

1. Calculate the distance AB .
2. State, with reasons but without any detailed calculations, what effect each of the following actions would have on the distance AB .

7 A projectile is fired from a point \(O\) on the slope of a hill which is inclined at an angle \(\alpha\) to the horizontal. The projectile is fired up the hill with velocity \(U\) at an angle \(\theta\) above the hill and first strikes it at a point \(A\). The projectile is modelled as a particle and the hill is modelled as a plane with \(O A\) as a line of greatest slope.

1. 1. Find, in terms of \(U , g , \alpha\) and \(\theta\), the time taken by the projectile to travel from \(O\) to \(A\).
   2. Hence, or otherwise, show that the magnitude of the component of the velocity of the projectile perpendicular to the hill, when it strikes the hill at the point \(A\), is the same as it was initially at \(O\).
2. The projectile rebounds and strikes the hill again at a point \(B\). The hill is smooth and the coefficient of restitution between the projectile and the hill is \(e\). \includegraphics[max width=\textwidth, alt={}, center]{fc5bfc4b-68bb-4a23-874b-87e9558dc990-06_428_1332_1023_338} Find the ratio of the time of flight from \(O\) to \(A\) to the time of flight from \(A\) to \(B\). Give your answer in its simplest form.

6 A football is kicked with speed \(31 \mathrm {~ms} ^ { - 1 }\) at an angle of \(20 ^ { \circ }\) to the horizontal. It travels towards the goal which is 50 m away. The height of the crossbar of the goal is 2.44 m .

1. Does the ball go over the top of the crossbar? Justify your answer.
2. State one assumption that you made in answering part (i).

6 A football is kicked with speed \(31 \mathrm {~ms} ^ { - 1 }\) at an angle of \(20 ^ { \circ }\) to the horizontal. It travels towards the goal which is 50 m away. The height of the crossbar of the goal is 2.44 m .

1. Does the ball go over the top of the crossbar? Justify your answer.
2. State one assumption that you made in answering part (i).

A particle \(P\) is projected with speed \(u\) at an angle \(\alpha\) above the horizontal from a point \(O\) on a horizontal plane and moves freely under gravity. The horizontal and vertical displacements of \(P\) from \(O\) at a subsequent time \(t\) are denoted by \(x\) and \(y\) respectively.

1. Derive the equation of the trajectory of \(P\) in the form $$y = x \tan \alpha - \frac{gx^2}{2u^2} \sec^2 \alpha.$$ [3]

During its flight, \(P\) must clear an obstacle of height \(h\) m that is at a horizontal distance of \(32\) m from the point of projection. When \(u = 40\sqrt{2}\) m s\(^{-1}\), \(P\) just clears the obstacle. When \(u = 40\) m s\(^{-1}\), \(P\) only achieves \(80\%\) of the height required to clear the obstacle.

1. Find the two possible values of \(h\). [6]

7 A particle is projected with a speed of \(8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), at an angle of \(40 ^ { \circ }\) above the horizontal. Find the speed and direction of motion of the particle 0.4 s after projection.

5 A particle \(P\) is projected with speed \(50 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of \(40 ^ { \circ }\) above the horizontal from a point \(O\). For the instant 2.5 s after projection, calculate

1. the speed of \(P\),
2. the angle between \(O P\) and the horizontal.

7. \begin{figure}[h] \end{figure} At time \(t = 0\), a particle \(P\) of mass 0.7 kg is projected with speed \(u \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from a fixed point \(O\) at an angle \(\theta ^ { \circ }\) to the horizontal. The particle moves freely under gravity. At time \(t = 2\) seconds, \(P\) passes through the point \(A\) with speed \(6 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and is moving downwards at \(45 ^ { \circ }\) to the horizontal, as shown in Figure 4. Find

1. the value of \(\theta\),
2. the kinetic energy of \(P\) as it reaches the highest point of its path. For an interval of \(T\) seconds, the speed, \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), of \(P\) is such that \(v \leqslant 6\)
3. Find the value of \(T\).

   VIIV STHI NI JINM ION OC

   VIAV SIHI NI JMAM/ION OC

   VIAV SIHI NI JIIYM ION OO

3 \includegraphics[max width=\textwidth, alt={}, center]{fe5c198d-5d05-4241-98f5-894ba92f7afe-3_408_1164_248_493} A particle \(P\) is released from rest at a point \(A\) which is 7 m above horizontal ground. At the same instant that \(P\) is released a particle \(Q\) is projected from a point \(O\) on the ground. The horizontal distance of \(O\) from \(A\) is 24 m . Particle \(Q\) moves in the vertical plane containing \(O\) and \(A\), with initial speed \(50 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and initial direction making an angle \(\theta\) above the horizontal, where \(\tan \theta = \frac { 7 } { 24 }\) (see diagram). Show that the particles collide.

5 \includegraphics[max width=\textwidth, alt={}, center]{36259e2a-aa9b-4655-b0c2-891f96c3f5a4-4_547_933_269_607} Particles \(A\) and \(B\) are projected simultaneously from the top \(T\) of a vertical tower, and move in the same vertical plane. \(T\) is 7.2 m above horizontal ground. \(A\) is projected horizontally with speed \(8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and \(B\) is projected at an angle of \(60 ^ { \circ }\) above the horizontal with speed \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 } . A\) and \(B\) move away from each other (see diagram).

1. Find the time taken for \(A\) to reach the ground. At the instant when \(A\) hits the ground,
2. show that \(B\) is approximately 5.2 m above the ground,
3. find the distance \(A B\).

A particle \(P\) is projected with speed \(32 \text{ m s}^{-1}\) at an angle of elevation \(\alpha\), where \(\sin \alpha = \frac{3}{4}\), from a point \(A\) on horizontal ground. At the same instant a particle \(Q\) is projected with speed \(20 \text{ m s}^{-1}\) at an angle of elevation \(\beta\), where \(\sin \beta = \frac{24}{25}\), from a point \(B\) on the same horizontal ground. The particles move freely under gravity in the same vertical plane and collide with each other at the point \(C\) at the instant when they are travelling horizontally (see diagram).

1. Calculate the height of \(C\) above the ground and the distance \(AB\). [4]

Immediately after the collision \(P\) falls vertically. \(P\) hits the ground and rebounds vertically upwards, coming to instantaneous rest at a height 5 m above the ground.

1. Given that the mass of \(P\) is 3 kg, find the magnitude and direction of the impulse exerted on \(P\) by the ground. [4]

The coefficient of restitution between the two particles is \(\frac{1}{2}\).

1. Find the distance of \(Q\) from \(C\) at the instant when \(Q\) is travelling in a direction of \(25°\) below the horizontal. [9]

A particle is projected with speed \(20 \text{ ms}^{-1}\) at an angle of \(60°\) above the horizontal. Calculate the time after projection when the particle is descending at an angle of \(40°\) below the horizontal. [4]

1 A particle \(P\) is projected with speed \(40 \mathrm {~ms} ^ { - 1 }\) at an angle of \(35 ^ { \circ }\) above the horizontal from a point \(O\). For the instant 3 s after projection, calculate the magnitude and direction of the velocity of \(P\).

A particle \(P\) is projected with speed \(u\) at an angle \(\theta\) above the horizontal from a point \(O\) on a horizontal plane and moves freely under gravity. The direction of motion of \(P\) makes an angle \(\alpha\) above the horizontal when \(P\) first reaches three-quarters of its greatest height.

1. Show that \(\tan \alpha = \frac{1}{2}\tan \theta\). [6]

In its subsequent motion, the greatest height reached by \(P\) above \(A\) is \(\frac{3}{10}\) of the vertical height of \(A\) above the horizontal plane.

1. Find the value of \(e\). [6]

A small ball is projected vertically downwards with speed \(5\text{ m s}^{-1}\) from a point \(A\) at a height of \(7.2\text{ m}\) above horizontal ground. The ball hits the ground with speed \(V\text{ m s}^{-1}\) and rebounds vertically upwards with speed \(\frac{1}{2}V\text{ m s}^{-1}\). The highest point the ball reaches after rebounding is \(B\). Find \(V\) and hence find the total time taken for the ball to reach the ground from \(A\) and rebound to \(B\). [5]

6 \begin{figure}[h] \end{figure} \begin{figure}[h] \end{figure} A small ball \(B\) is projected with speed \(U \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of \(\theta ^ { \circ }\) above the horizontal from a point \(O\). At time 2 s after the instant of projection, \(B\) strikes a smooth wall which slopes at \(60 ^ { \circ }\) to the horizontal. The speed of \(B\) is \(18 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and its direction of motion is perpendicular to the wall at the instant of impact (see Fig. 1). \(B\) bounces off the wall with speed \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in a direction perpendicular to the wall. At time 0.8 s after \(B\) bounces off the wall, \(B\) strikes the wall again at a lower point \(A\) (see Fig. 2).

1. Find \(U\) and \(\theta\).
2. By considering the motion of \(B\) after it bounces off the wall, calculate \(V\).

9 A particle is projected from a point \(O\) on horizontal ground with speed \(40 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle \(\theta\) above the horizontal.

1. Write down the equation of the trajectory, in terms of \(\tan \theta\).
2. The particle passes through a point whose horizontal and vertical distances from \(O\) are 72 m and \(y \mathrm {~m}\) respectively. By considering the equation of the trajectory as a quadratic equation in \(\tan \theta\), or otherwise, find the greatest possible value of \(y\).

5 A particle is projected from a point \(O\) on horizontal ground. The velocity of projection has magnitude \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and direction upwards at an angle \(\theta\) to the horizontal. The particle passes through the point which is 7 m above the ground and 16 m horizontally from \(O\), and hits the ground at the point \(A\).

1. Using the equation of the particle's trajectory and the identity \(\sec ^ { 2 } \theta = 1 + \tan ^ { 2 } \theta\), show that the possible values of \(\tan \theta\) are \(\frac { 3 } { 4 }\) and \(\frac { 17 } { 4 }\).
2. Find the distance \(O A\) for each of the two possible values of \(\tan \theta\).
3. Sketch in the same diagram the two possible trajectories.

Fig. 12 shows \(x\)- and \(y\)- coordinate axes with origin O and the trajectory of a particle projected from O with speed 28 m s\(^{-1}\) at an angle \(\alpha\) to the horizontal. After \(t\) seconds, the particle has horizontal and vertical displacements \(x\) m and \(y\) m. Air resistance should be neglected. \includegraphics{figure_12}

1. Show that the equation of the trajectory is given by $$\tan^2\alpha - \frac{160}{x}\tan\alpha + \frac{160y}{x^2} + 1 = 0.$$ (*) [5]
2. 1. Show that if (*) is treated as an equation with \(\tan\alpha\) as a variable and with \(x\) and \(y\) as constants, then (*) has two distinct real roots for \(\tan\alpha\) when \(y < 40 - \frac{x^2}{160}\). [3]
   2. Show the inequality in part (ii)(A) as a locus on the graph of \(y = 40 - \frac{x^2}{160}\) in the Printed Answer Booklet and label it R. [1]

S is the locus of points \((x, y)\) where (*) has one real root for \(\tan\alpha\). T is the locus of points \((x, y)\) where (*) has no real roots for \(\tan\alpha\).

1. Indicate S and T on the graph in the Printed Answer Booklet. [2]
2. State the significance of R, S and T for the possible trajectories of the particle. [3]

A machine can fire a tennis ball from ground level with a maximum speed of 28 m s\(^{-1}\).

1. State, with a reason, whether a tennis ball fired from the machine can achieve a range of 80 m. [1]

A particle \(P\) is projected vertically upwards with speed \(24 \text{ m s}^{-1}\) from a point \(5 \text{ m}\) above ground level. Find the time from projection until \(P\) reaches the ground. [3]

A particle \(P\) is projected vertically upwards with speed \(24 \text{ m s}^{-1}\) from a point \(5 \text{ m}\) above ground level. Find the time from projection until \(P\) reaches the ground. [3]

A particle \(P\) is projected from a point \(O\) on a horizontal plane and moves freely under gravity. The initial velocity of \(P\) is 100 ms\(^{-1}\) at an angle \(\theta\) above the horizontal, where \(\tan\theta = \frac{4}{3}\). The two times at which \(P\)'s height above the plane is \(H\) m differ by 10 s.

1. Find the value of \(H\). [5]

1. Find the magnitude and direction of the velocity of \(P\) one second before it strikes the plane. [4]

1 A golf ball \(B\) is projected from a point \(O\) on horizontal ground. \(B\) hits the ground for the first time at a point 48 m away from \(O\) at time 2.4 s after projection. Calculate the angle of projection.

1 A golf ball \(B\) is projected from a point \(O\) on horizontal ground. \(B\) hits the ground for the first time at a point 48 m away from \(O\) at time 2.4 s after projection. Calculate the angle of projection.

2 A small ball is projected from a point 1.5 m above horizontal ground. At a point 9 m above the ground the ball is travelling at \(45 ^ { \circ }\) above the horizontal and its velocity is \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Find the angle of projection of the ball.

\includegraphics{figure_1} A smooth solid hemisphere, of radius 0.8 m and centre \(O\), is fixed with its plane face on a horizontal table. A particle of mass 0.5 kg is projected horizontally with speed \(u\) m s\(^{-1}\) from the highest point \(A\) of the hemisphere. The particle leaves the hemisphere at the point \(B\), which is a vertical distance of 0.2 m below the level of \(A\). The speed of the particle at \(B\) is \(v\) m s\(^{-1}\) and the angle between \(OA\) and \(OB\) is \(\theta\), as shown in Fig. 1.

1. Find the value of \(\cos \theta\). [1]
2. Show that \(v^2 = 5.88\). [3]
3. Find the value of \(u\). [3]

A ball is projected forward from a fixed point, \(P\), on a horizontal surface with an initial speed \(u\text{ ms}^{-1}\), at an acute angle \(\theta\) above the horizontal. The ball needs to first land at a point at least \(d\) metres away from \(P\). You may assume the ball may be modelled as a particle and that air resistance may be ignored. Show that $$\sin 2\theta \geq \frac{dg}{u^2}$$ [6 marks]

At time \(t = 0\) seconds, a particle \(P\) is projected with speed \(u\) m s\(^{-1}\) at an angle \(60°\) above the horizontal from a point \(O\). In the subsequent motion \(P\) moves freely under gravity. The direction of motion of \(P\) when \(t = 5\) is perpendicular to its direction of motion when \(t = 15\). Find the value of \(u\). [5]

A particle \(P\) is projected with speed \(20 \text{ m s}^{-1}\) at an angle of \(60°\) below the horizontal, from a point \(O\) which is \(30 \text{ m}\) above horizontal ground.

1. Calculate the time taken by \(P\) to reach the ground. [3]
2. Calculate the speed and direction of motion of \(P\) immediately before it reaches the ground. [4]

12 A train of mass 250 tonnes is ascending an incline of \(\sin ^ { - 1 } \left( \frac { 1 } { 500 } \right)\) and working at 400 kW against resistance to motion which may be regarded as a constant force of 20000 N .

1. Find the constant speed, \(V\), with which the train can ascend the incline working at this power.
2. The train begins to ascend the incline at \(6 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at the same power and against the same resistance. Find the distance covered in reaching a speed of \(\frac { 3 } { 4 } V\).

10 Ship \(A\) is 15 km due south of ship \(B\). Ship \(B\) is travelling at \(20 \mathrm {~km} \mathrm {~h} ^ { - 1 }\) on a bearing of \(300 ^ { \circ }\). Ship \(A\) is travelling at \(16 \mathrm {~km} \mathrm {~h} ^ { - 1 }\). Find

1. the bearing, to the nearest degree, that \(A\) must take in order to get as close as possible to \(B\), [4]
2. the time, in minutes, that it takes for the ships to be as close as possible.

A particle is projected vertically upwards from a point \(O\) with initial speed \(12.5 \text{ m s}^{-1}\). At the same instant another particle is released from rest at a point 10 m vertically above \(O\). Find the height above \(O\) at which the particles meet. [5]