3.02i Projectile motion: constant acceleration model

6 A small ball is kicked off the edge of a jetty over a calm sea. Air resistance is negligible. Fig. 6 shows

• the point of projection, O,
• the initial horizontal and vertical components of velocity,
• the point A on the jetty vertically below O and at sea level,
• the height, OA , of the jetty above the sea.

\begin{figure}[h] \end{figure} The time elapsed after the ball is kicked is \(t\) seconds.

1. Find an expression in terms of \(t\) for the height of the ball above O at time \(t\). Find also an expression for the horizontal distance of the ball from O at this time.
2. Determine how far the ball lands from A .

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.

2 A ball is kicked from ground level over horizontal ground. It leaves the ground at a speed of 25 ms 1 and at an angle \(\theta\) to the horizontal such that \(\cos \theta = 0.96\) and \(\sin \theta = 0.28\).

1. Show that the height, \(y \mathrm {~m}\), of the ball above the ground \(t\) seconds after projection is given by \(y = 7 t - 4.9 t ^ { 2 }\). Show also that the horizontal distance, \(x \mathrm {~m}\), travelled by this time is given by \(x = 24 t\).
2. Calculate the maximum height reached by the ball.
3. Calculate the times at which the ball is at half its maximum height. Find the horizontal distance travelled by the ball between these times.
4. Determine the following when \(t = 1.25\).
   (A) The vertical component of the velocity of the ball.
   (B) Whether the ball is rising or falling. (You should give a reason for your answer.)
   (C) The speed of the ball.
5. Show that the equation of the trajectory of the ball is $$y = \frac { 0.7 x } { 576 } ( 240 - 7 x )$$ Hence, or otherwise, find the range of the ball.

4 You should neglect air resistance in this question.
A small stone is projected from ground level. The maximum height of the stone above horizontal ground is 22.5 m .

1. Show that the vertical component of the initial velocity of the stone is \(21 \mathrm {~ms} { } ^ { 1 }\). The speed of projection is \(28 \mathrm {~ms} { } ^ { 1 }\).
2. Find the angle of projection of the stone.
3. Find the horizontal range of the stone.

5 In this question take the value of \(\boldsymbol { g }\) to be \(\mathbf { 1 0 ~ } \mathbf { m ~ s } ^ { \mathbf { 2 } }\). \(\Lambda\) particle \(\Lambda\) is projected over horizontal ground from a point P which is 9 m above a point O on the ground. The initial velocity has horizontal and vertical components of \(10 \mathrm {~ms} ^ { - 1 }\) and \(12 \mathrm {~ms} ^ { - 1 }\) respectively, as shown in Fig. 7. The trajectory of the particle meets the ground at X. Air resistance may be neglected. \begin{figure}[h] \end{figure}

1. Calculate the specd of projection \(u \mathrm {~ms} ^ { - 1 }\) and the angle of projection \(\theta ^ { \circ }\).
2. Show that, \(t\) seconds after projection, the height of particle A above the ground is \(9 + 12 t - 5 t ^ { 2 }\). Write down an expression in terms of \(t\) for the horizontal distance of the particle from O at this time.
3. Calculate the maximum height of particle \(\Lambda\) above the point of projection.
4. Calculate the distance OX . \(\wedge\) second particle, \(B\), is projected from \(O\) with speed \(20 \mathrm {~ms} ^ { - 1 }\) at \(60 ^ { \circ }\) to the horizontal. The trajectories of A and B are in the same vertical plane. Particles A and B are projected at the same time.
5. Show that the horizontal displacements of A and B are always cqual.
6. Show that, \(t\) seconds after projection, the height of particle B above the ground is \(10 \sqrt { 3 } t - 5 t ^ { 2 }\).
7. Show that the particles collide 1.7 seconds after projection (correct to two significant figures).

8. A particle \(P\) is projected from a point \(O\) with initial velocity \(( 3 \cdot 5 \mathbf { i } + 12 \mathbf { j } ) \mathrm { ms } ^ { - 1 }\) and moves under gravity. \(\mathbf { i }\) and \(\mathbf { j }\) are unit vectors in the horizontal and vertical directions respectively.

1. Find the initial speed of \(P\).
2. Show that the position vector \(\mathbf { r } \mathbf { m }\) of \(P\) at time \(t\) seconds after projection is given by $$\mathbf { r } = 3 \cdot 5 t \mathbf { i } + \left( 12 t - 4 \cdot 9 t ^ { 2 } \right) \mathbf { j } .$$
3. Find the horizontal distance of \(P\) from \(O\) at each of the times when it is 4.4 m vertically above the level of \(O\). In a refined model of the motion of \(P\), the position vector of \(P\) at time \(t\) seconds is taken to be $$\mathbf { r } = 3 \cdot 5 t \mathbf { i } + \left( 12 t - t ^ { 3 } \right) \mathbf { j } \mathbf { ~ m } .$$
4. Using this model, find the position vector of the highest point reached by \(P\).

A ball is hit with initial speed \(u \mathrm {~ms} ^ { - 1 }\), at an angle \(\theta\) above the horizontal, from a point at a height of \(h \mathrm {~m}\) above horizontal ground. The ball, which is modelled as a particle moving freely under gravity, hits the ground at a horizontal distance \(d \mathrm {~m}\) from the point of projection.

1. Prove that \(\frac { g d ^ { 2 } } { 2 u ^ { 2 } } \sec ^ { 2 } \theta - d \tan \theta - h = 0\). Given further that \(u = 14 , h = 7\) and \(d = 14\), and assuming the result \(\sec ^ { 2 } \theta = 1 + \tan ^ { 2 } \theta\),
2. find the value of \(\theta\).
   

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 \(27 ^ { \circ }\).

4 A ball is projected from a point \(O\) on the edge of a vertical cliff. The horizontal and vertically upward components of the initial velocity are \(7 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and \(21 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) respectively. At time \(t\) seconds after projection the ball is at the point \(( x , y )\) referred to horizontal and vertically upward axes through \(O\). Air resistance may be neglected.

1. Express \(x\) and \(y\) in terms of \(t\), and hence show that \(y = 3 x - \frac { 1 } { 10 } x ^ { 2 }\). The ball hits the sea at a point which is 25 m below the level of \(O\).
2. Find the horizontal distance between the cliff and the point where the ball hits the sea.

4 A golfer hits a ball from a point \(O\) on horizontal ground with a velocity of \(35 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of \(\theta\) above the horizontal. The horizontal range of the ball is \(R\) metres and the time of flight is \(t\) seconds.

1. Express \(t\) in terms of \(\theta\), and hence show that \(R = 125 \sin 2 \theta\). The golfer hits the ball so that it lands 110 m from \(O\).
2. Calculate the two possible values of \(t\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{6ae57fe9-3b6f-46c2-95b8-d48903ed796b-3_672_403_267_872} \captionsetup{labelformat=empty} \caption{Fig. 1}
   \end{figure} A toy is constructed by attaching a small ball of mass 0.01 kg to one end of a uniform rod of length 10 cm whose other end is attached to the centre of the plane face of a uniform solid hemisphere with radius 3 cm . The rod has mass 0.02 kg , the hemisphere has mass 0.5 kg and the rod is perpendicular to the plane face of the hemisphere (see Fig. 1).

7 \includegraphics[max width=\textwidth, alt={}, center]{6ae57fe9-3b6f-46c2-95b8-d48903ed796b-4_305_1301_1708_424} Two small spheres \(A\) and \(B\) of masses 2 kg and 3 kg respectively lie at rest on a smooth horizontal platform which is fixed at a height of 4 m above horizontal ground (see diagram). Sphere \(A\) is given an impulse of 6 N s towards \(B\), and \(A\) then strikes \(B\) directly. The coefficient of restitution between \(A\) and \(B\) is \(\frac { 2 } { 3 }\).

1. Show that the speed of \(B\) after it has been hit by \(A\) is \(2 \mathrm {~ms} ^ { - 1 }\). Sphere \(B\) leaves the platform and follows the path of a projectile.
2. Calculate the speed and direction of motion of \(B\) at the instant when it hits the ground.

7 \includegraphics[max width=\textwidth, alt={}, center]{e85c2bf4-21a8-4d9a-93c5-d5679b2a8233-4_440_657_906_744} A ball is projected with an initial speed of \(17 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of \(25 ^ { \circ }\) below the horizontal from a point on the top of a vertical wall. The point of projection is 8 m above horizontal ground. The ball hits a vertical fence which is at a horizontal distance of 9 m from the wall (see diagram).

1. Calculate the height above the ground of the point where the ball hits the fence.
2. Calculate the direction of motion of the ball immediately before it hits the fence.
3. It is given that \(30 \%\) of the kinetic energy of the ball is lost when it hits the fence. Calculate the speed of the ball immediately after it hits the fence.

5 A particle is projected with speed \(7 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of elevation of \(30 ^ { \circ }\) from a point \(O\) and moves freely under gravity. The horizontal and vertically upwards displacements of the particle from \(O\) at any subsequent time \(t \mathrm {~s}\) are \(x \mathrm {~m}\) and \(y \mathrm {~m}\) respectively.

1. Express \(x\) and \(y\) in terms of \(t\) and hence find the equation of the trajectory of the particle.
2. Calculate the values of \(x\) when \(y = 0.6\).
3. Find the direction of motion of the particle when \(y = 0.6\) and the particle is rising.

4 A boy throws a small ball at a vertical wall. The ball is thrown horizontally, from a point \(O\), at a speed of \(14.4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and it hits the wall at a point which is 0.2 m below the level of \(O\).

1. Find the horizontal distance from \(O\) to the wall. The boy now moves so that he is 6 m from the wall. He throws the ball at an angle of \(15 ^ { \circ }\) above the horizontal. The ball again hits the wall at a point which is 0.2 m below the level from which it was thrown.
2. Find the speed at which the ball was thrown.

6 \includegraphics[max width=\textwidth, alt={}, center]{74eaa61a-1507-4cef-8f97-5c1860bdc36a-4_243_1179_1580_443} The masses of two particles \(A\) and \(B\) are 0.2 kg and \(m \mathrm {~kg}\) respectively. The particles are moving with constant speeds \(4 \mathrm {~ms} ^ { - 1 }\) and \(u \mathrm {~ms} ^ { - 1 }\) in the same horizontal line and in the same direction (see diagram). The two particles collide and the coefficient of restitution between the particles is \(e\). After the collision, \(A\) and \(B\) continue in the same direction with speeds \(4 \left( 1 - e + e ^ { 2 } \right) \mathrm { ms } ^ { - 1 }\) and \(4 \mathrm {~ms} ^ { - 1 }\) respectively.

1. Find \(u\) and \(m\) in terms of \(e\).
2. Find the value of \(e\) for which the speed of \(A\) after the collision is least and find, in this case, the total loss in kinetic energy due to the collision.
3. Find the possible values of \(e\) for which the magnitude of the impulse that \(B\) exerts on \(A\) is 0.192 Ns . \includegraphics[max width=\textwidth, alt={}, center]{74eaa61a-1507-4cef-8f97-5c1860bdc36a-5_744_887_264_589} The diagram shows a surface consisting of a horizontal part \(O A\) and a plane \(A B\) inclined at an angle of \(70 ^ { \circ }\) to the horizontal. A particle is projected from the point \(O\) with speed \(u \mathrm {~ms} ^ { - 1 }\) at an angle of \(\theta ^ { \circ }\) above the horizontal \(O A\). The particle hits the plane \(A B\) at the point \(P\), with speed \(14 \mathrm {~ms} ^ { - 1 }\) and at right angles to the plane, 1.4 s after projection.

1. Show that the value of \(u\) is 15.9 , correct to 3 significant figures, and find the value of \(\theta\).
2. Find the height of \(P\) above the level of \(A\). The particle rebounds with speed \(v \mathrm {~ms} ^ { - 1 }\). The particle next lands at \(A\).
3. Find the value of \(v\).
4. Find the coefficient of restitution between the particle and the plane at \(P\).

1 A football is kicked from horizontal ground with speed \(20 \mathrm {~ms} ^ { - 1 }\) at an angle of \(\theta ^ { \circ }\) above the horizontal. The greatest height the football reaches above ground level is 2.44 m . By modelling the football as a particle and ignoring air resistance, find

1. the value of \(\theta\),
2. the range of the football.

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 \(V\).

6 A particle is projected with speed \(v \mathrm {~ms} ^ { - 1 }\) from a point \(O\) on horizontal ground. The angle of projection is \(\theta ^ { \circ }\) above the horizontal. At time \(t\) seconds after the instant of projection the horizontal displacement of the particle from \(O\) is \(x \mathrm {~m}\) and the upward vertical displacement from \(O\) is \(y \mathrm {~m}\).

1. Show that $$y = x \tan \theta - \frac { 4.9 x ^ { 2 } } { v ^ { 2 } \cos ^ { 2 } \theta } .$$ A stone is thrown from the top of a vertical cliff 100 m high. The initial speed of the stone is \(16 \mathrm {~ms} ^ { - 1 }\) and the angle of projection is \(\theta ^ { \circ }\) to the horizontal. The stone hits the sea 40 m from the foot of the cliff.
2. Find the two possible values of \(\theta\). \includegraphics[max width=\textwidth, alt={}, center]{8492ec9b-3327-4d89-aaa4-bf98cdf0ebdc-3_623_995_1475_536} A uniform ladder \(A B\) of weight \(W \mathrm {~N}\) and length 4 m rests with its end \(A\) on rough horizontal ground and its end \(B\) against a smooth vertical wall. The ladder is inclined at an angle \(\theta\) to the horizontal where \(\tan \theta = \frac { 1 } { 2 }\) (see diagram). A small object \(S\) of weight \(2 W \mathrm {~N}\) is placed on the ladder at a point \(C\), which is 1 m from \(A\). The coefficient of friction between the ladder and the ground is \(\mu\) and the system is in limiting equilibrium.

6 An athlete 'puts the shot' with an initial speed of \(19 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of \(11 ^ { \circ }\) above the horizontal. At the instant of release the shot is 1.53 m above the horizontal ground. By treating the shot as a particle and ignoring air resistance, find

1. the maximum height, above the ground, reached by the shot,
2. the horizontal distance the shot has travelled when it hits the ground.

1

1. Sphere P , of mass 10 kg , and sphere Q , of mass 15 kg , move with their centres on a horizontal straight line and have no resistances to their motion. \(\mathrm { P } , \mathrm { Q }\) and the positive direction are shown in Fig. 1.1. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{1dd32b82-020e-45ef-8146-892197fd0985-2_332_803_434_712} \captionsetup{labelformat=empty} \caption{Fig. 1.1}
   \end{figure} Initially, P has a velocity of \(- 1.75 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and is acted on by a force of magnitude 13 N acting in the direction PQ . After \(T\) seconds, P has a velocity of \(4.75 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and has not reached Q .
   1. Calculate \(T\). The force of magnitude 13 N is removed. P is still travelling at \(4.75 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) when it collides directly with Q , which has a velocity of \(- 0.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Suppose that P and Q coalesce in the collision to form a single object.
   2. Calculate their common velocity after the collision. Suppose instead that P and Q separate after the collision and that P has a velocity of \(1 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) afterwards.
   3. Calculate the velocity of Q after the collision and also the coefficient of restitution in the collision.
2. Fig. 1.2 shows a small ball projected at a speed of \(14 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of \(30 ^ { \circ }\) below the horizontal over smooth horizontal ground. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{1dd32b82-020e-45ef-8146-892197fd0985-2_424_832_1918_699} \captionsetup{labelformat=empty} \caption{Fig. 1.2}
   \end{figure} The ball is initially 3.125 m above the ground. The coefficient of restitution between the ball and the ground is 0.6 . Calculate the angle with the horizontal of the ball's trajectory immediately after the second bounce on the ground.

1

1. In this part-question, all the objects move along the same straight line on a smooth horizontal plane. All their collisions are direct. The masses of the objects \(\mathrm { P } , \mathrm { Q }\) and R and the initial velocities of P and Q (but not R ) are shown in Fig. 1.1. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{c8f26b7e-1be1-4abf-8fea-6847185fad81-2_177_1011_488_529} \captionsetup{labelformat=empty} \caption{Fig. 1.1}
   \end{figure} A force of 21 N acts on P for 2 seconds in the direction \(\mathrm { PQ } . \mathrm { P }\) does not reach Q in this time.
   1. Calculate the speed of P after the 2 seconds. The force of 21 N is removed after the 2 seconds. When P collides with Q they stick together (coalesce) to form an object S of mass 6 kg .
   2. Show that immediately after the collision S has a velocity of \(8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) towards R . The collision between S and R is elastic with coefficient of restitution \(\frac { 1 } { 4 }\). After the collision, S has a velocity of \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in the direction of its motion before the collision.
   3. Find the velocities of R before and after the collision. \section*{(b) In this part-question take \(\boldsymbol { g } = \mathbf { 1 0 }\).} A particle of mass 0.2 kg is projected vertically downwards with initial speed \(5 \mathrm {~ms} ^ { - 1 }\) and it travels 10 m before colliding with a fixed smooth plane. The plane is inclined at \(\alpha\) to the vertical where \(\tan \alpha = \frac { 3 } { 4 }\). Immediately after its collision with the plane, the particle has a speed of \(13 \mathrm {~ms} ^ { - 1 }\). This information is shown in Fig. 1.2. Air resistance is negligible. \begin{figure}[h]
      \includegraphics[alt={},max width=\textwidth]{c8f26b7e-1be1-4abf-8fea-6847185fad81-2_383_341_1795_854} \captionsetup{labelformat=empty} \caption{Fig. 1.2}
      \end{figure}
   4. Calculate the angle between the direction of motion of the particle and the plane immediately after the collision. Calculate also the coefficient of restitution in the collision.
   5. Calculate the magnitude of the impulse of the plane on the particle.

4

1. Two discs, P of mass 4 kg and Q of mass 5 kg , are sliding along the same line on a smooth horizontal plane when they collide. The velocity of P before the collision and the velocity of Q after the collision are shown in Fig. 4. P loses \(\frac { 5 } { 9 }\) of its kinetic energy in the collision. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{71d839d8-12ca-4806-8f74-c572e7e21891-5_294_899_390_584} \captionsetup{labelformat=empty} \caption{Fig. 4}
   \end{figure}
   1. Show that after the collision P has a velocity of \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in the opposite direction to its original motion. While colliding, the discs are in contact for \(\frac { 1 } { 5 } \mathrm {~s}\).
   2. Find the impulse on P in the collision and the average force acting on the discs.
   3. Find the velocity of Q before the collision and the coefficient of restitution between the two discs.
2. A particle is projected from a point 2.5 m above a smooth horizontal plane. Its initial velocity is \(5.95 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle \(\theta\) below the horizontal, where \(\sin \theta = \frac { 15 } { 17 }\). The coefficient of restitution between the particle and the plane is \(\frac { 4 } { 5 }\).
   1. Show that, after bouncing off the plane, the greatest height reached by the particle is 2.5 m .
   2. Calculate the horizontal distance between the two points at which the particle is 2.5 m above the plane.

7. \begin{figure}[h] \end{figure} Figure 3 shows the path of a golf ball which is hit from the point \(O\) with speed \(49 \mathrm {~ms} ^ { - 1 }\) at an angle of \(30 ^ { \circ }\) to the horizontal. The path of the ball is in a vertical plane containing \(O\) and the hole at which the ball is aimed. The hole is 170 m from \(O\) and on the same horizontal level as \(O\).

1. Suggest a suitable model for the motion of the golf ball. Find, correct to 3 significant figures,
2. the distance beyond the hole at which the ball hits the ground,
3. the magnitude and direction of the velocity of the ball when it is directly above the hole.