Projection from elevated point - angle above horizontal

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.

6. \begin{figure}[h] \end{figure} A small ball \(P\) is projected with speed \(7 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from a point \(A 10 \mathrm {~m}\) above horizontal ground. The angle of projection is \(55 ^ { \circ }\) above the horizontal. The ball moves freely under gravity and hits the ground at the point \(B\), as shown in Figure 3. Find

1. the speed of \(P\) as it hits the ground at \(B\),
2. the direction of motion of \(P\) as it hits the ground at \(B\),
3. the time taken for \(P\) to move from \(A\) to \(B\).

4 Fig. 4 illustrates a situation in which a film is being made. A cannon is fired from the top of a vertical cliff towards a ship out at sea. The director wants the cannon ball to fall just short of the ship so that it appears to be a near-miss. There are actors on the ship so it is important that it is not hit by mistake. The cannon ball is fired from a height 75 m above the sea with an initial velocity of \(20 \mathrm {~ms} ^ { - 1 }\) at an angle of \(30 ^ { \circ }\) above the horizontal. The ship is 90 m from the bottom of the cliff. \begin{figure}[h] \end{figure}

1. The director calculates where the cannon ball will hit the sea, using the standard projectile model and taking the value of \(g\) to be \(10 \mathrm {~ms} ^ { - 2 }\). Verify that according to this model the cannon ball is in the air for 5 seconds. Show that it hits the water less than 5 m from the ship.
2. Without doing any further calculations state, with a brief reason, whether the cannon ball would be predicted to travel further from the cliff if the value of \(g\) were taken to be \(9.8 \mathrm {~ms} ^ { - 2 }\).

7. \begin{figure}[h] \end{figure} A ball is projected with speed \(40 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from a point \(P\) on a cliff above horizontal ground. The point O on the ground is vertically below P and OP is 36 m . The ball is projected at an angle \(\theta ^ { \circ }\) to the horizontal. The point Q is the highest point of the path of the ball and is 12 m above the level of P. The ball moves freely under gravity and hits the ground at the point R , as shown in Figure 3. Find

1. the value of \(\theta\),
2. the distance OR ,
3. the speed of the ball as it hits the ground at R.

7. \begin{figure}[h] \end{figure} A particle \(P\) is projected from a point \(A\) with speed \(32 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of elevation \(\alpha\), where \(\sin \alpha = \frac { 3 } { 5 }\). The point \(O\) is on horizontal ground, with \(O\) vertically below \(A\) and \(O A = 20 \mathrm {~m}\). The particle \(P\) moves freely under gravity and passes through a point \(B\), which is 16 m above ground, before reaching the ground at the point \(C\), as shown in Figure 4. Calculate

1. the time of the flight from \(A\) to \(C\),
2. the distance \(O C\),
3. the speed of \(P\) at \(B\),
4. the angle that the velocity of \(P\) at \(B\) makes with the horizontal.

6. \begin{figure}[h] \end{figure} A ball is thrown from a point \(O\), which is 6 m above horizontal ground. The ball is projected with speed \(u \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle \(\theta\) above the horizontal. There is a thin vertical post which is 4 m high and 8 m horizontally away from the vertical through \(O\), as shown in Figure 2. The ball passes just above the top of the post 2 s after projection. The ball is modelled as a particle.

1. Show that \(\tan \theta = 2.2\)
2. Find the value of \(u\). The ball hits the ground \(T\) seconds after projection.
3. Find the value of \(T\). Immediately before the ball hits the ground the direction of motion of the ball makes an angle \(\alpha\) with the horizontal.
4. Find \(\alpha\).

6. \begin{figure}[h] \end{figure} A golf ball \(P\) is projected with speed \(35 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from a point \(A\) on a cliff above horizontal ground. The angle of projection is \(\alpha\) to the horizontal, where \(\tan \alpha = \frac { 4 } { 3 }\). The ball moves freely under gravity and hits the ground at the point \(B\), as shown in Figure 4.

1. Find the greatest height of \(P\) above the level of \(A\). The horizontal distance from \(A\) to \(B\) is 168 m .
2. Find the height of \(A\) above the ground. By considering energy, or otherwise,
3. find the speed of \(P\) as it hits the ground at \(B\).

7. \begin{figure}[h] \end{figure} A shot is projected upwards from the top of a cliff with a velocity of \(28 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of \(30 ^ { \circ }\) above the horizontal. It strikes the ground 52.5 m vertically below the level of the point of projection, as shown in Fig. 3. The motion of the shot is modelled as that of a particle moving freely under gravity. Find, to 3 significant figures,

1. the horizontal distance from the point of projection at which the shot strikes the ground,
2. the speed of the shot as it strikes the ground.

7 A small ball is projected at an angle of \(50 ^ { \circ }\) above the horizontal, from a point \(A\), which is 2 m above ground level. The highest point of the path of the ball is 15 m above the ground, which is horizontal. Air resistance may be ignored.

1. Find the speed with which the ball is projected from \(A\). The ball hits a net at a point \(B\) when it has travelled a horizontal distance of 45 m .
2. Find the height of \(B\) above the ground.
3. Find the speed of the ball immediately before it hits the net.

5. \begin{figure}[h] \end{figure} A small ball is projected with speed \(U \mathrm {~ms} ^ { - 1 }\) from a point \(O\) at the top of a vertical cliff. The point \(O\) is 25 m vertically above the point \(N\) which is on horizontal ground. The ball is projected at an angle of \(45 ^ { \circ }\) above the horizontal.
The ball hits the ground at a point \(A\), where \(A N = 100 \mathrm {~m}\), as shown in Figure 2 .
The motion of the ball is modelled as that of a particle moving freely under gravity.
Using this initial model,

1. show that \(U = 28\)
2. find the greatest height of the ball above the horizontal ground \(N A\). In a refinement to the model of the motion of the ball from \(O\) to \(A\), the effect of air resistance is included. This refined model is used to find a new value of \(U\).
3. How would this new value of \(U\) compare with 28, the value given in part (a)?
4. State one further refinement to the model that would make the model more realistic. \section*{" " \(_ { \text {" } } ^ { \text {" } }\) " "}

4. \begin{figure}[h] \end{figure} A small stone is projected with speed \(65 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from a point \(O\) at the top of a vertical cliff.
Point \(O\) is 70 m vertically above the point \(N\).
Point \(N\) is on horizontal ground.
The stone is projected at an angle \(\alpha\) above the horizontal, where \(\tan \alpha = \frac { 5 } { 12 }\) The stone hits the ground at the point \(A\), as shown in Figure 3.
The stone is modelled as a particle moving freely under gravity.
The acceleration due to gravity is modelled as having magnitude \(\mathbf { 1 0 m ~ s } \mathbf { m ~ } ^ { \mathbf { - 2 } }\) Using the model,

1. find the time taken for the stone to travel from \(O\) to \(A\),
2. find the speed of the stone at the instant just before it hits the ground at \(A\). One limitation of the model is that it ignores air resistance.
3. State one other limitation of the model that could affect the reliability of your answers.

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}

6 Emma is in a park with her dog, Roxy. Emma throws a ball and Roxy catches it in her mouth. The ground in the park is horizontal. Emma throws the ball from a point at a height of 1.2 metres above the ground and Roxy catches the ball when it is at a height of 0.5 metres above the ground. Emma throws the ball with an initial velocity of \(8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of \(30 ^ { \circ }\) above the horizontal.

1. Find the time that the ball takes to travel from Emma's hand to Roxy's mouth.
2. Find the horizontal distance travelled by the ball during its flight.
3. During the flight, the speed of the ball is a maximum when it is at a height of \(h\) metres above the ground. Write down the value of \(h\).
4. Find the maximum speed of the ball during its flight.
   [0pt] [4 marks]
   \includegraphics[max width=\textwidth, alt={}]{01338c87-302c-420f-a473-39b0796ccaed-14_1566_1707_1137_153}

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).

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.

5. \begin{figure}[h] \end{figure} During a cricket match, a batsman hits the ball giving it an initial velocity of \(22 \mathrm {~ms} ^ { - 1 }\) at an angle \(\alpha\) to the horizontal where \(\sin \alpha = \frac { 7 } { 8 }\). When the batsman strikes the ball it is 1.6 metres above the ground, as shown in Figure 2, and it subsequently moves freely under gravity.

1. Find, correct to 3 significant figures, the maximum height above the ground reached by the ball. The ball is caught by a fielder when it is 0.2 metres above the ground.
2. Find the length of time for which the ball is in the air. Assuming that the fielder who caught the ball ran at a constant speed of \(6 \mathrm {~m} \mathrm {~s} ^ { - 1 }\),
3. find, correct to 3 significant figures, the maximum distance that the fielder could have been from the ball when it was struck.

10. \begin{figure}[h] \end{figure} A boy throws a stone with speed \(U \mathrm {~ms} ^ { - 1 }\) from a point \(O\) at the top of a vertical cliff. The point \(O\) is 18 m above sea level.
The stone is thrown at an angle \(\alpha\) above the horizontal, where \(\tan \alpha = \frac { 3 } { 4 }\).
The stone hits the sea at the point \(S\) which is at a horizontal distance of 36 m from the foot of the cliff, as shown in Figure 2.
The stone is modelled as a particle moving freely under gravity with \(g = 10 \mathrm {~m} \mathrm {~s} ^ { - 2 }\) Find

1. the value of \(U\),
2. the speed of the stone when it is 10.8 m above sea level, giving your answer to 2 significant figures.
3. Suggest two improvements that could be made to the model.

11 A particle \(P\) of mass 2 kg can move along a line of greatest slope on the smooth surface of a wedge which is fixed to the ground. The sloping face \(O A\) of the wedge has length 10 metres and is inclined at \(30 ^ { \circ }\) to the horizontal (see Fig. 1). \(P\) is fired up the slope from the lowest point \(O\), with an initial speed of \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). \begin{figure}[h] \end{figure}

1. Find the time taken for \(P\) to reach \(A\) and show that the speed of \(P\) at \(A\) is \(10 \sqrt { 3 } \mathrm {~m} \mathrm {~s} ^ { - 1 }\). After \(P\) has reached \(A\) it becomes a projectile (see Fig. 2). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{f0c32e07-f3a0-4d58-bd00-c266177ceaac-5_424_1533_1123_306} \captionsetup{labelformat=empty} \caption{Fig. 2}
   \end{figure}
2. Find the total horizontal distance travelled by \(P\) from \(O\) when it hits the ground.

\includegraphics{figure_4} A small ball \(B\) is projected from a point \(O\) above horizontal ground, with initial speed \(15\) m s\(^{-1}\) at an angle of projection of \(30°\) above the horizontal (see diagram). The ball strikes the ground \(3\) s after projection.

1. Calculate the speed and direction of motion of the ball immediately before it strikes the ground. [5]
2. Find the height of \(O\) above the ground. [2]

\includegraphics{figure_2} At time \(t = 0\) a small package is projected from a point \(B\) which is \(2.4\) m above a point \(A\) on horizontal ground. The package is projected with speed \(23.75\) m s\(^{-1}\) at an angle \(α\) to the horizontal, where \(\tan α = \frac{4}{5}\). The package strikes the ground at point \(C\), as shown in Fig. 2. The package is modelled as a particle moving freely under gravity.

1. Find the time taken for the package to reach \(C\). [5]

A lorry moves along the line \(AC\), approaching \(A\) with constant speed 18 m s\(^{-1}\). At time \(t = 0\) the rear of the lorry passes \(A\) and the lorry starts to slow down. It comes to rest 7 seconds later. The acceleration, \(a\) m s\(^{-2}\) of the lorry at time \(t\) seconds is given by $$a = -\frac{1}{4}t^2, \quad 0 \leq t \leq 7.$$

1. Find the speed of the lorry at time \(t\) seconds. [3]

1. Hence show that \(T = 6\). [3]

1. Show that when the package reaches \(C\) it is just under 10 m behind the rear of the moving lorry. [5]

END

\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

\includegraphics{figure_2} At time \(t = 0\) a small package is projected from a point \(B\) which is 2.4 m above a point \(A\) on horizontal ground. The package is projected with speed 23.75 m s\(^{-1}\) at an angle \(\alpha\) to the horizontal, where \(\tan \alpha = \frac{4}{3}\). The package strikes the ground at the point \(C\), as shown in Fig. 2. The package is modelled as a particle moving freely under gravity.

1. Find the time taken for the package to reach \(C\). [5]

A lorry moves along the line \(AC\), approaching \(A\) with constant speed 18 m s\(^{-1}\). At time \(t = 0\) the rear of the lorry passes \(A\) and the lorry starts to slow down. It comes to rest \(T\) seconds later. The acceleration, \(a\) m s\(^{-2}\) of the lorry at time \(t\) seconds is given by $$a = -\frac{1}{4}t^2, \quad 0 \leq t \leq T.$$

1. Find the speed of the lorry at time \(t\) seconds. [3]
2. Hence show that \(T = 6\). [3]
3. Show that when the package reaches \(C\) it is just under 10 m behind the rear of the moving lorry. [5]

END

\includegraphics{figure_3} A ball is thrown from a point 4 m above horizontal ground. The ball is projected at an angle \(\alpha\) above the horizontal, where \(\tan \alpha = \frac{4}{3}\). The ball hits the ground at a point which is a horizontal distance 8 m from its point of projection, as shown in Fig. 3. The initial speed of the ball is \(u\) m s\(^{-1}\) and the time of flight is \(T\) seconds.

1. Prove that \(uT = 10\). [2]
2. Find the value of \(u\). [5]

As the ball hits the ground, its direction of motion makes an angle \(\phi\) with the horizontal.

1. Find \(\tan \phi\). [5]

\includegraphics{figure_3} A ball is projected with speed 40 m s\(^{-1}\) from a point \(P\) on a cliff above horizontal ground. The point \(O\) on the ground is vertically below \(P\) and \(OP\) is 36 m. The ball is projected at an angle \(\theta°\) to the horizontal. The point \(Q\) is the highest point of the path of the ball and is 12 m above the level of \(P\). The ball moves freely under gravity and hits the ground at the point \(R\), as shown in Figure 3. Find

1. the value of \(\theta\), (3)
2. the distance \(OR\), (6)
3. the speed of the ball as it hits the ground at \(R\). (3)