Projectile passing through given point

7 A ball is projected from a point \(O\) with speed \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\), at an angle of \(30 ^ { \circ }\) above the horizontal. At time \(T\) s after projection, the ball passes through the point \(A\), whose horizontal and vertically upward displacements from \(O\) are 10 m and 2 m respectively.

1. By using the equation of the trajectory, or otherwise, find the value of \(V\).
2. Find the value of \(T\).
3. Find the angle that the direction of motion of the ball at \(A\) makes with the horizontal.

6 A particle is projected from a point \(O\) on horizontal ground. The velocity of projection has magnitude \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and direction upwards at \(35 ^ { \circ }\) to the horizontal. The particle passes through the point \(M\) at time \(T\) seconds after the instant of projection. The point \(M\) is 2 m above the ground and at a horizontal distance of 25 m from \(O\).

1. Find the values of \(V\) and \(T\).
2. Find the speed of the particle as it passes through \(M\) and determine whether it is moving upwards or downwards.

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 is projected from a point \(O\) at an angle of \(35 ^ { \circ }\) above the horizontal. At time \(T\) s later the particle passes through a point \(A\) whose horizontal and vertically upward displacements from \(O\) are 8 m and 3 m respectively.

1. By using the equation of the particle's trajectory, or otherwise, find (in either order) the speed of projection of the particle from \(O\) and the value of \(T\).
2. Find the angle between the direction of motion of the particle at \(A\) and the horizontal. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{b9080e9f-2c23-43ce-b171-bd68648dc56b-5_476_895_269_625} \captionsetup{labelformat=empty} \caption{Fig. 1}
   \end{figure} Fig. 1 shows the cross-section of a uniform solid. The cross-section has the shape and dimensions shown. The centre of mass \(C\) of the solid lies in the plane of this cross-section. The distance of \(C\) from \(D E\) is \(y \mathrm {~cm}\).
3. Find the value of \(y\). The solid is placed on a rough plane. The coefficient of friction between the solid and the plane is \(\mu\). The plane is tilted so that \(E F\) lies along a line of greatest slope.
4. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{b9080e9f-2c23-43ce-b171-bd68648dc56b-5_375_431_1366_897} \captionsetup{labelformat=empty} \caption{Fig. 2}
   \end{figure} The solid is placed so that \(F\) is higher up the plane than \(E\) (see Fig. 2). When the angle of inclination is sufficiently great the solid starts to topple (without sliding). Show that \(\mu > \frac { 1 } { 2 }\). [3]
5. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{b9080e9f-2c23-43ce-b171-bd68648dc56b-5_376_428_2069_900} \captionsetup{labelformat=empty} \caption{Fig. 3}
   \end{figure} The solid is now placed so that \(E\) is higher up the plane than \(F\) (see Fig. 3). When the angle of inclination is sufficiently great the solid starts to slide (without toppling). Show that \(\mu < \frac { 5 } { 6 }\). [3]

7 A particle \(P\) is projected from a point \(O\) on horizontal ground with speed \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and direction \(60 ^ { \circ }\) upwards from the horizontal. At time \(t \mathrm {~s}\) later the horizontal and vertical displacements of \(P\) from \(O\) are \(x \mathrm {~m}\) and \(y \mathrm {~m}\) respectively.

1. Write down expressions for \(x\) and \(y\) in terms of \(V\) and \(t\) and hence show that the equation of the trajectory of \(P\) is $$y = ( \sqrt { } 3 ) x - \frac { 20 x ^ { 2 } } { V ^ { 2 } }$$ \(P\) passes through the point \(A\) at which \(x = 70\) and \(y = 10\). Find
2. the value of \(V\),
3. the direction of motion of \(P\) at the instant it passes through \(A\).

5 A particle \(P\) is projected with speed \(u \mathrm {~ms} ^ { - 1 }\) at an angle of \(\theta\) 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 \(t \mathrm {~s}\) are denoted by \(x \mathrm {~m}\) and \(y \mathrm {~m}\) respectively.

1. Starting from the equation of the trajectory given in the List of formulae (MF19), show that $$\mathrm { y } = \mathrm { x } \tan \theta - \frac { \mathrm { gx } ^ { 2 } } { 2 \mathrm { u } ^ { 2 } } \left( 1 + \tan ^ { 2 } \theta \right)$$ When \(\theta = \tan ^ { - 1 } 2 , P\) passes through the point with coordinates \(( 10,16 )\).
2. Show that there is no value of \(\theta\) for which \(P\) can pass through the point with coordinates \(( 18,30 )\).

7 \includegraphics[max width=\textwidth, alt={}, center]{af7d1fc8-5552-48b8-a359-895b2b5d3d6c-4_433_841_255_653} A particle \(P\) is projected from a point \(O\) with initial speed \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of \(45 ^ { \circ }\) above the horizontal. \(P\) subsequently passes through the point \(A\) which is at an angle of elevation of \(30 ^ { \circ }\) from \(O\) (see diagram). At time \(t \mathrm {~s}\) after projection the horizontal and vertically upward displacements of \(P\) from \(O\) are \(x \mathrm {~m}\) and \(y \mathrm {~m}\) respectively.

1. Write down expressions for \(x\) and \(y\) in terms of \(t\), and hence obtain the equation of the trajectory of \(P\).
2. Calculate the value of \(x\) when \(P\) is at \(A\).
3. Find the angle the trajectory makes with the horizontal when \(P\) is at \(A\).

2 \includegraphics[max width=\textwidth, alt={}, center]{d1f1f036-1676-443e-b733-ca1fe79972d4-2_525_913_1123_616} A particle \(P\) is projected from a point \(O\) at an angle of \(60 ^ { \circ }\) above horizontal ground. At an instant 0.6 s after projection, the angle of elevation of \(P\) from \(O\) is \(45 ^ { \circ }\) (see diagram).

1. Show that the speed of projection of \(P\) is \(8.20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), correct to 3 significant figures.
2. Calculate the time after projection when the direction of motion of \(P\) is \(45 ^ { \circ }\) above the horizontal.

3 A particle \(P\) is projected with speed \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of \(\theta ^ { \circ }\) above the horizontal from a point \(O\) on horizontal ground. At the instant 4 s after projection the particle passes through the point \(A\), where \(O A = 40 \mathrm {~m}\) and the line \(O A\) makes an angle of \(30 ^ { \circ }\) with the horizontal. Calculate \(V\) and \(\theta\).

3 A particle \(P\) is projected with speed \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of \(\theta ^ { \circ }\) above the horizontal from a point \(O\) on horizontal ground. At the instant 4 s after projection the particle passes through the point \(A\), where \(O A = 40 \mathrm {~m}\) and the line \(O A\) makes an angle of \(30 ^ { \circ }\) with the horizontal. Calculate \(V\) and \(\theta\).

6. \begin{figure}[h] \end{figure} A child playing cricket on horizontal ground hits the ball towards a fence 10 m away. The ball moves in a vertical plane which is perpendicular to the fence. The ball just passes over the top of the fence, which is 2 m above the ground, as shown in Figure 3. The ball is modelled as a particle projected with initial speed \(u \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from point \(O\) on the ground at an angle \(\alpha\) to the ground.

1. By writing down expressions for the horizontal and vertical distances, from \(O\) of the ball \(t\) seconds after it was hit, show that $$2 = 10 \tan \alpha - \frac { 50 g } { u ^ { 2 } \cos ^ { 2 } \alpha }$$ Given that \(\alpha = 45 ^ { \circ }\),
2. find the speed of the ball as it passes over the fence.

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 \(\beta - \alpha\)

7 In this question take \(\boldsymbol { g } = \mathbf { 1 0 }\).
A small stone is projected from a point O with a speed of \(26 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle \(\theta\) above the horizontal. The initial velocity and part of the path of the stone are shown in Fig. 7.
You are given that \(\sin \theta = \frac { 12 } { 13 }\).
After \(t\) seconds the horizontal displacement of the stone from O is \(x\) metres and the vertical displacement is \(y\) metres. \begin{figure}[h] \end{figure}

1. Using the standard model for projectile motion,
   The stone passes through a point A . Point A is 16 m above the level of O .
2. Find the two possible horizontal distances of A from O . A toy balloon is projected from O with the same initial velocity as the small stone.
3. Suggest two ways in which the standard model could be adapted.

8 A golf ball is hit from a point on a horizontal surface, so that it has an initial velocity \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle \(\alpha\) above the horizontal. The ball travels through the air and after 2.4 seconds hits a vertical wall at a height of 3 metres. The wall is at a horizontal distance of 38.4 metres from the point where the ball was hit. The path of the ball is shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{ccc1db66-9700-4f22-905e-cc0bdf1fd3c1-18_300_1000_566_520} Assume that the weight of the ball is the only force that acts on it as it travels through the air.

1. Find the horizontal component of the velocity of the ball.
2. \(\quad\) Find \(V\).
3. \(\quad\) Find \(\alpha\).

6 A bullet is fired from a rifle at a target, which is at a distance of 420 metres from the rifle. The bullet leaves the rifle travelling at \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and at an angle of \(2 ^ { \circ }\) above the horizontal. The centre of the target, \(C\), is at the same horizontal level as the rifle. The bullet hits the target at the point \(A\), which is on a vertical line through \(C\). The bullet takes 1.8 seconds to reach the point \(A\).

1. Find \(V\), showing clearly how you obtain your answer.
2. Find the distance between \(A\) and \(C\).
3. State one assumption that you have made about the forces acting on the bullet.
   [0pt] [1 mark]

7 Fig. 4 shows a particle projected over horizontal ground from a point O at ground level. The particle initially has a speed of \(32 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle \(\alpha\) to the horizontal. The particle is a horizontal distance of 44.8 m from O after 5 seconds. Air resistance should be neglected. \begin{figure}[h] \end{figure}

1. Write down an expression, in terms of \(\alpha\) and \(t\), for the horizontal distance of the particle from O at time \(t\) seconds after it is projected.
2. Show that \(\cos \alpha = 0.28\).
3. Calculate the greatest height reached by the particle.

8 A child is trying to throw a small stone to hit a target painted on a vertical wall. The child and the wall are on horizontal ground. The child is standing a horizontal distance of 8 m from the base of the wall. The child throws the stone from a height of 1 m with speed \(12 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of \(20 ^ { \circ }\) above the horizontal.

1. Find the direction of motion of the stone when it hits the wall. The child now throws the stone with a speed of \(\mathrm { Vm } \mathrm { s } ^ { - 1 }\) from the same initial position and still at an angle of \(20 ^ { \circ }\) above the horizontal. This time the stone hits the target which is 2.5 m above the ground.
2. Find \(V\).

5 A ball is projected with speed \(u \mathrm {~ms} ^ { - 1 }\) at an angle of elevation \(\alpha\) above the horizontal so as to hit a point \(P\) on a wall. The ball travels in a vertical plane through the point of projection. During the motion, the horizontal and upward vertical displacements of the ball from the point of projection are \(x\) metres and \(y\) metres respectively.

1. Show that, during the flight, the equation of the trajectory of the ball is given by $$y = x \tan \alpha - \frac { g x ^ { 2 } } { 2 u ^ { 2 } } \left( 1 + \tan ^ { 2 } \alpha \right)$$
2. The ball is projected from a point 1 metre vertically below and \(R\) metres horizontally from the point \(P\).
   1. By taking \(g = 10 \mathrm {~ms} ^ { - 2 }\), show that \(R\) satisfies the equation $$5 R ^ { 2 } \tan ^ { 2 } \alpha - u ^ { 2 } R \tan \alpha + 5 R ^ { 2 } + u ^ { 2 } = 0$$
   2. Hence, given that \(u\) and \(R\) are constants, show that, for \(\tan \alpha\) to have real values, \(R\) must satisfy the inequality $$R ^ { 2 } \leqslant \frac { u ^ { 2 } \left( u ^ { 2 } - 20 \right) } { 100 }$$
   3. Given that \(R = 5\), determine the minimum possible speed of projection.

5 A boy throws a small ball from a height of 1.5 m above horizontal ground with initial velocity \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle \(\alpha\) above the horizontal. The ball hits a small can placed on a vertical wall of height 2.5 m , which is at a horizontal distance of 5 m from the initial position of the ball, as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{eed9842d-cd89-4eb7-b5ba-9380971be196-3_499_1180_1283_424}

1. Show that \(\alpha\) satisfies the equation $$49 \tan ^ { 2 } \alpha - 200 \tan \alpha + 89 = 0$$
2. Find the two possible values of \(\alpha\), giving your answers to the nearest \(0.1 ^ { \circ }\).
3. 1. To knock the can off the wall, the horizontal component of the velocity of the ball must be greater than \(8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Show that, for one of the possible values of \(\alpha\) found in part (b), the can will be knocked off the wall, and for the other, it will not be knocked off the wall.
      (3 marks)
   2. Given that the can is knocked off the wall, find the direction in which the ball is moving as it hits the can.

2 A projectile is fired from a point \(O\) on top of a hill with initial velocity \(80 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle \(\theta\) above the horizontal and moves in a vertical plane. The horizontal and upward vertical distances of the projectile from \(O\) are \(x\) metres and \(y\) metres respectively.

1. 1. Show that, during the flight, the equation of the trajectory of the projectile is given by $$y = x \tan \theta - \frac { g x ^ { 2 } } { 12800 } \left( 1 + \tan ^ { 2 } \theta \right)$$
   2. The projectile hits a target \(A\), which is 20 m vertically below \(O\) and 400 m horizontally from \(O\). \includegraphics[max width=\textwidth, alt={}, center]{01071eb0-2c48-4028-8cd3-6021ce86d7e5-04_392_1031_970_460} Show that $$49 \tan ^ { 2 } \theta - 160 \tan \theta + 41 = 0$$
2. 1. Find the two possible values of \(\theta\). Give your answers to the nearest \(0.1 ^ { \circ }\).
   2. Hence find the shortest possible time of the flight of the projectile from \(O\) to \(A\).
3. State a necessary modelling assumption for answering part (a)(i).
   \includegraphics[max width=\textwidth, alt={}]{01071eb0-2c48-4028-8cd3-6021ce86d7e5-05_2484_1709_223_153}
   \includegraphics[max width=\textwidth, alt={}]{01071eb0-2c48-4028-8cd3-6021ce86d7e5-07_2484_1709_223_153}

3 (In this question, use \(g = 10 \mathrm {~m} \mathrm {~s} ^ { - 2 }\).)
A golf ball is hit from a point \(O\) on a horizontal golf course with a velocity of \(40 \mathrm {~ms} ^ { - 1 }\) at an angle of elevation \(\theta\). The golf ball travels in a vertical plane through \(O\). During its flight, the horizontal and upward vertical distances of the golf ball from \(O\) are \(x\) and \(y\) metres respectively.

1. Show that the equation of the trajectory of the golf ball during its flight is given by $$x ^ { 2 } \tan ^ { 2 } \theta - 320 x \tan \theta + \left( x ^ { 2 } + 320 y \right) = 0$$
2. 1. The golf ball hits the top of a tree, which has a vertical height of 8 m and is at a horizontal distance of 150 m from \(O\). Find the two possible values of \(\theta\).
   2. Which value of \(\theta\) gives the shortest possible time for the golf ball to travel from \(O\) to the top of the tree? Give a reason for your choice of \(\theta\).

3 (In this question, take \(g = 10 \mathrm {~m} \mathrm {~s} ^ { - 2 }\).)
A projectile is fired from a point \(O\) with speed \(u\) at an angle of elevation \(\alpha\) above the horizontal so as to pass through a point \(P\). The projectile travels in a vertical plane through \(O\) and \(P\). The point \(P\) is at a horizontal distance \(2 k\) from \(O\) and at a vertical distance \(k\) above \(O\).

1. Show that \(\alpha\) satisfies the equation $$20 k \tan ^ { 2 } \alpha - 2 u ^ { 2 } \tan \alpha + u ^ { 2 } + 20 k = 0$$
2. Deduce that $$u ^ { 4 } - 20 k u ^ { 2 } - 400 k ^ { 2 } \geqslant 0$$

3 A player projects a basketball with speed \(u \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle \(\theta\) above the horizontal. The basketball travels in a vertical plane through the point of projection and goes into the basket. During the motion, the horizontal and upward vertical displacements of the basketball from the point of projection are \(x\) metres and \(y\) metres respectively. \includegraphics[max width=\textwidth, alt={}, center]{3a1726d9-1b0c-41de-8b43-56019e18aac1-06_737_937_513_550}

1. Find an expression for \(y\) in terms of \(x , u , g\) and \(\tan \theta\).
2. The player projects the basketball with speed \(8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from a point 0.5 metres vertically below and 5 metres horizontally from the basket.
   1. Show that the two possible values of \(\theta\) are approximately \(63.1 ^ { \circ }\) and \(32.6 ^ { \circ }\), correct to three significant figures.
   2. Given that the player projects the basketball at \(63.1 ^ { \circ }\) to the horizontal, find the direction of the motion of the basketball as it enters the basket. Give your answer to the nearest degree.
3. State a modelling assumption needed for answering parts (a) and (b) of this question.
   (1 mark)

12 A girl is practising netball.
She throws the ball from a height of 1.5 m above horizontal ground and aims to get the ball through a hoop.
The hoop is 2.5 m vertically above the ground and is 6 m horizontally from the point of projection. The situation is modelled as follows.

• The initial velocity of the ball has magnitude \(U \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
• The angle of projection is \(40 ^ { \circ }\).
• The ball is modelled as a particle.
• The hoop is modelled as a point.

This is shown on the diagram below. \includegraphics[max width=\textwidth, alt={}, center]{ec83c2c5-f8f8-4357-abfa-d40bc1d026b4-09_375_1207_1119_278}

1. For \(U = 10\), find
   1. the greatest height above the ground reached by the ball
   2. the distance between the ball and the hoop when the ball is vertically above the hoop.
2. Calculate the value of \(U\) which allows her to hit the hoop.
3. How appropriate is this model for predicting the path of the ball when it is thrown by the girl?
4. Suggest one improvement that might be made to this model.

10. \begin{figure}[h] \end{figure} A boy throws a ball at a target. At the instant when the ball leaves the boy's hand at the point \(A\), the ball is 2 m above horizontal ground and is moving with speed \(U\) at an angle \(\alpha\) above the horizontal. In the subsequent motion, the highest point reached by the ball is 3 m above the ground. The target is modelled as being the point \(T\), as shown in Figure 4.
The ball is modelled as a particle moving freely under gravity.
Using the model,

1. show that \(U ^ { 2 } = \frac { 2 g } { \sin ^ { 2 } \alpha }\). The point \(T\) is at a horizontal distance of 20 m from \(A\) and is at a height of 0.75 m above the ground. The ball reaches \(T\) without hitting the ground.
2. Find the size of the angle \(\alpha\)
3. State one limitation of the model that could affect your answer to part (b).
4. Find the time taken for the ball to travel from \(A\) to \(T\).