3.02i Projectile motion: constant acceleration model

\includegraphics{figure_3} [In this question, the unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are in a vertical plane, \(\mathbf{i}\) being horizontal and \(\mathbf{j}\) being vertical.] A particle \(P\) is projected from the point \(A\) which has position vector \(47.5\mathbf{j}\) metres with respect to a fixed origin \(O\). The velocity of projection of \(P\) is \((2u\mathbf{i} + 5u\mathbf{j})\) m s\(^{-1}\). The particle moves freely under gravity passing through the point \(B\) with position vector \(30\mathbf{i}\) metres, as shown in Figure 3.

1. Show that the time taken for \(P\) to move from \(A\) to \(B\) is 5 s. [6]
2. Find the value of \(u\). [2]
3. Find the speed of \(P\) at \(B\). [5]

[In this question \(\mathbf{i}\) and \(\mathbf{j}\) are unit vectors in a horizontal and upward vertical direction respectively] A particle \(P\) is projected from a fixed point \(O\) on horizontal ground with velocity \(u(\mathbf{i} + c\mathbf{j}) \text{ ms}^{-1}\), where \(c\) and \(u\) are positive constants. The particle moves freely under gravity until it strikes the ground at \(A\), where it immediately comes to rest. Relative to \(O\), the position vector of a point on the path of \(P\) is \((x\mathbf{i} + y\mathbf{j})\) m.

1. Show that $$y = cx - \frac{4.9x^2}{u^2}.$$ [5]

Given that \(u = 7\), \(OA = R\) m and the maximum vertical height of \(P\) above the ground is \(H\) m,

1. using the result in part (a), or otherwise, find, in terms of \(c\),
   1. \(R\)
   2. \(H\).
   [6]

Given also that when \(P\) is at the point \(Q\), the velocity of \(P\) is at right angles to its initial velocity,

1. find, in terms of \(c\), the value of \(x\) at \(Q\). [6]

[In this question, the unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are horizontal and vertical respectively.] \includegraphics{figure_3} The point \(O\) is a fixed point on a horizontal plane. A ball is projected from \(O\) with velocity \((6\mathbf{i} + 12\mathbf{j})\) m s\(^{-1}\), and passes through the point \(A\) at time \(t\) seconds after projection. The point \(B\) is on the horizontal plane vertically below \(A\), as shown in Figure 3. It is given that \(OB = 2AB\). Find

1. the value of \(t\), [7]
2. the speed, \(V\) m s\(^{-1}\), of the ball at the instant when it passes through \(A\). [5]

At another point \(C\) on the path the speed of the ball is also \(V\) m s\(^{-1}\).

1. Find the time taken for the ball to travel from \(O\) to \(C\). [3]

The unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) lie in a vertical plane, \(\mathbf{i}\) being horizontal and \(\mathbf{j}\) vertical. A ball of mass 0.1 kg is hit by a bat which gives it an impulse of \((3.5\mathbf{i} + 3\mathbf{j})\) Ns. The velocity of the ball immediately after being hit is \((10\mathbf{i} + 25\mathbf{j})\) m s\(^{-1}\).

1. Find the velocity of the ball immediately before it is hit. [3]

In the subsequent motion the ball is modelled as a particle moving freely under gravity. When it is hit the ball is 1 m above horizontal ground.

1. Find the greatest height of the ball above the ground in the subsequent motion. [3]

The ball is caught when it is again 1 m above the ground.

1. Find the distance from the point where the ball is hit to the point where it is caught. [4]

\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

A particle is projected from a point with speed \(u\) at an angle of elevation \(\alpha\) above the horizontal and moves freely under gravity. When it has moved a horizontal distance \(x\), its height above the point of projection is \(y\).

1. Show that $$y = x \tan \alpha - \frac{gx^2}{2u^2}(1 + \tan^2 \alpha).$$ [5]

A shot-putter puts a shot from a point \(A\) at a height of 2 m above horizontal ground. The shot is projected at an angle of elevation of 45° with a speed of 14 m s\(^{-1}\). By modelling the shot as a particle moving freely under gravity,

1. find, to 3 significant figures, the horizontal distance of the shot from \(A\) when the shot hits the ground, [5]
2. find, to 2 significant figures, the time taken by the shot in moving from \(A\) to reach the ground. [2]

\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]

A vertical cliff is 73.5 m high. Two stones \(A\) and \(B\) are projected simultaneously. Stone \(A\) is projected horizontally from the top of the cliff with speed 28 m s\(^{-1}\). Stone \(B\) is projected from the bottom of the cliff with speed 35 m s\(^{-1}\) at an angle \(\alpha\) above the horizontal. The stones move freely under gravity in the same vertical plane and collide in mid-air. By considering the horizontal motion of each stone,

1. prove that \(\cos \alpha = \frac{4}{5}\). [4]

1. Find the time which elapses between the instant when the stones are projected and the instant when they collide. [4]

\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)

A particle is projected from a point \(O\) with speed \(u\) at an angle of elevation \(\alpha\) above the horizontal and moves freely under gravity. When the particle has moved a horizontal distance \(x\), its height above \(O\) is \(y\).

1. Show that $$y = x \tan \alpha - \frac{gx^2}{2u^2 \cos^2 \alpha}$$ [4]

A girl throws a ball from a point \(A\) at the top of a cliff. The point \(A\) is 8 m above a horizontal beach. The ball is projected with speed 7 m s\(^{-1}\) at an angle of elevation of 45°. By modelling the ball as a particle moving freely under gravity,

1. find the horizontal distance of the ball from \(A\) when the ball is 1 m above the beach. [5]

A boy is standing on the beach at the point \(B\) vertically below \(A\). He starts to run in a straight line with speed \(v\) m s\(^{-1}\), leaving \(B\) 0.4 seconds after the ball is thrown. He catches the ball when it is 1 m above the beach.

1. Find the value of \(v\). [4]

\includegraphics{figure_4} A small ball is projected from a fixed point \(O\) so as to hit a target \(T\) which is at a horizontal distance \(9a\) from \(O\) and at a height \(6a\) above the level of \(O\). The ball is projected with speed \(\sqrt{(27ag)}\) at an angle \(\theta\) to the horizontal, as shown in Figure 4. The ball is modelled as a particle moving freely under gravity.

1. Show that tan\(^2 \theta - 6\) tan \(\theta + 5 = 0\) [7]

The two possible angles of projection are \(\theta_1\) and \(\theta_2\), where \(\theta_1 > \theta_2\).

1. Find tan \(\theta_1\) and tan \(\theta_2\). [3]

The particle is projected at the larger angle \(\theta_1\).

1. Show that the time of flight from \(O\) to \(T\) is \(\sqrt{\left(\frac{78a}{g}\right)}\). [3]
2. Find the speed of the particle immediately before it hits \(T\). [3]

\includegraphics{figure_1} The points \(O\) and \(B\) are on horizontal ground. The point \(A\) is \(h\) metres vertically above \(O\). A particle \(P\) is projected from \(A\) with speed 12 m s\(^{-1}\) at an angle \(\alpha°\) to the horizontal. The particle moves freely under gravity and hits the ground at \(B\), as shown in Figure 1. The speed of \(P\) immediately before it hits the ground is 15 m s\(^{-1}\).

1. By considering energy, find the value of \(h\). [4]

Given that 1.5 s after it is projected from \(A\), \(P\) is at a point 4 m above the level of \(A\), find

1. the value of \(\alpha\), [3]
2. the direction of motion of \(P\) immediately before it reaches \(B\). [3]

\includegraphics{figure_1} A smooth solid hemisphere of radius 0.5 m is fixed with its plane face on a horizontal floor. The plane face has centre \(O\) and the highest point of the surface of the hemisphere is \(A\). A particle \(P\) has mass 0.2 kg. The particle is projected horizontally with speed \(u\) m s\(^{-1}\) from \(A\) and leaves the hemisphere at the point \(B\), where \(OB\) makes an angle \(\theta\) with \(OA\), as shown in Figure 1. The point \(B\) is at a vertical distance of 0.1 m below the level of \(A\). The speed of \(P\) at \(B\) is \(v\) m s\(^{-1}\)

1. Show that \(v^2 = u^2 + 1.96\) [3]
2. Find the value of \(u\). [4]

The particle first strikes the floor at the point \(C\).

1. Find the length of \(OC\). [7]

\includegraphics{figure_3} Figure 3 shows a fixed hollow sphere of internal radius \(a\) and centre \(O\). A particle \(P\) of mass \(m\) is projected horizontally from the lowest point \(A\) of a sphere with speed \(\sqrt{\left(\frac{5}{4}ag\right)}\). It moves in a vertical circle, centre \(O\), on the smooth inner surface of the sphere. The particle passes through the point \(B\), which is in the same horizontal plane as \(O\). It leaves the surface of the sphere at the point \(C\), where \(OC\) makes an angle \(\theta\) with the upward vertical.

1. Find, in terms of \(m\) and \(g\), the normal reaction between \(P\) and the surface of the sphere at \(B\). [4]
2. Show that \(\theta = 60°\). [7]

After leaving the surface of the sphere, \(P\) meets it again at the point \(A\).

1. Find, in terms of \(a\) and \(g\), the time \(P\) takes to travel from \(C\) to \(A\). [4]

One end of a light inextensible string of length \(l\) is attached to a fixed point \(A\). The other end is attached to a particle \(P\) of mass \(m\), which is held at a point \(B\) with the string taut and \(AP\) making an angle arccos \(\frac{1}{4}\) with the downward vertical. The particle is released from rest. When \(AP\) makes an angle \(\theta\) with the downward vertical, the string is taut and the tension in the string is \(T\).

1. Show that $$T = 3mg \cos \theta - \frac{mg}{2}.$$ [6]

\includegraphics{figure_3} At an instant when \(AP\) makes an angle of \(60°\) to the downward vertical, \(P\) is moving upwards, as shown in Figure 3. At this instant the string breaks. At the highest point reached in the subsequent motion, \(P\) is at a distance \(d\) below the horizontal through \(A\).

1. Find \(d\) in terms of \(l\). [5]

\includegraphics{figure_2} A smooth sphere of radius \(a\) is fixed with a point \(A\) of its surface in contact with a fixed vertical wall. A particle is placed on the highest point of the sphere and is projected towards the wall and perpendicular to the wall with horizontal speed \(\sqrt{\frac{2ag}{5}}\), as shown in Figure 2. The particle leaves the surface of the sphere with speed \(V\).

1. Show that \(V = \sqrt{\frac{4ag}{5}}\) [7]

The particle strikes the wall at the point \(X\).

1. Find the distance \(AX\). [9]

Two stones are projected simultaneously from a point \(O\) on horizontal ground. Stone \(A\) is thrown vertically upwards with speed \(98\) ms\(^{-1}\). Stone \(B\) is projected along the smooth ground in a straight line at \(24.5\) ms\(^{-1}\).

1. Find the distances of the two stones from \(O\) after \(t\) seconds, where \(0 \leq t \leq 20\). \hfill [3 marks]
2. Show that the distance \(d\) m between the two stones after \(t\) seconds is given by $$d^2 = 24.01(t^2 - 40t^2 + 425t^2).$$ \hfill [6 marks]
3. Hence find the range of values of \(t\) for which the distance between the stones is decreasing. \hfill [6 marks]

A small firework is fired from a point O at ground level over horizontal ground. The highest point reached by the firework is a horizontal distance of 60 m from O and a vertical distance of 40 m from O, as shown in Fig. 7. Air resistance is negligible. The initial horizontal component of the velocity of the firework is 21 m s\(^{-1}\).

1. Calculate the time for the firework to reach its highest point and show that the initial vertical component of its velocity is 28 m s\(^{-1}\). [4]
2. Show that the firework is \((28t - 4.9t^2)\) m above the ground \(t\) seconds after its projection. [1]

When the firework is at its highest point it explodes into several parts. Two of the parts initially continue to travel horizontally in the original direction, one with the original horizontal speed of 21 m s\(^{-1}\) and the other with a quarter of this speed.

1. State why the two parts are always at the same height as one another above the ground and hence find an expression in terms of \(t\) for the distance between the parts \(t\) seconds after the explosion. [3]
2. Find the distance between these parts of the firework
   1. when they reach the ground, [2]
   2. when they are 10 m above the ground. [5]
3. Show that the cartesian equation of the trajectory of the firework before it explodes is \(y = \frac{4}{90}(120x - x^2)\), referred to the coordinate axes shown in Fig. 7. [4]

Anila is practising catching tennis balls. She uses a mobile computer-controlled machine which fires tennis balls vertically upwards from a height of 2.5 metres above the ground. Once it has fired a ball, the machine is programmed to move position rapidly to allow Anila time to get into a suitable position to catch the ball. The machine fires a ball at 24 ms\(^{-1}\) vertically upwards and Anila catches the ball just before it touches the ground.

1. Draw a speed-time graph for the motion of the ball from the time it is fired by the machine to the instant before Anila catches it. [3 marks]
2. Find, to the nearest centimetre, the maximum height which the ball reaches above the ground. [4 marks]
3. Calculate the speed at which the ball is travelling when Anila catches it. [4 marks]
4. Calculate the length of time that the ball is in the air. [3 marks]

Fig. 7 shows the trajectory of an object which is projected from a point O on horizontal ground. Its initial velocity is \(40\text{ms}^{-1}\) at an angle of \(\alpha\) to the horizontal. \includegraphics{figure_1}

1. Show that, according to the standard projectile model in which air resistance is neglected, the flight time, \(T\) s, and the range, \(R\) m, are given by $$T = \frac{80\sin\alpha}{g} \text{ and } R = \frac{3200\sin\alpha\cos\alpha}{g}.$$ [6] A company is designing a new type of ball and wants to model its flight.
2. Initially the company uses the standard projectile model. Use this model to show that when \(\alpha = 30°\) and the initial speed is \(40\text{ms}^{-1}\), \(T\) is approximately \(4.08\) and \(R\) is approximately \(141.4\). Find the values of \(T\) and \(R\) when \(\alpha = 45°\). [3] The company tests the ball using a machine that projects it from ground level across horizontal ground. The speed of projection is set at \(40\text{ms}^{-1}\). When the angle of projection is set at \(30°\), the range is found to be \(125\) m.
3. Comment briefly on the accuracy of the standard projectile model in this situation. [1] The company refines the model by assuming that the ball has a constant deceleration of \(2\text{ms}^{-2}\) in the horizontal direction. In this new model, the resistance to the vertical motion is still neglected and so the flight time is still \(4.08\) s when the angle of projection is \(30°\).
4. Using the new model, with \(\alpha = 30°\), show that the horizontal displacement from the point of projection, \(x\) m at time \(t\) s, is given by $$x = 40t\cos 30° - t^2.$$ Find the range and hence show that this new model is reasonably accurate in this case. [4] The company then sets the angle of projection to \(45°\) while retaining a projection speed of \(40\text{ms}^{-1}\). With this setting the range of the ball is found to be \(135\) m.
5. Investigate whether the new model is also accurate for this angle of projection. [3]
6. Make one suggestion as to how the model could be further refined. [1]

\includegraphics{figure_2} Fig. 7 shows a platform \(10\) m long and \(2\) m high standing on horizontal ground. A small ball projected from the surface of the platform at one end, O, just misses the other end, P. The ball is projected at \(68.5°\) to the horizontal with a speed of \(U\text{ms}^{-1}\). Air resistance may be neglected. At time \(t\) seconds after projection, the horizontal and vertical displacements of the ball from O are \(x\) m and \(y\) m.

1. Obtain expressions, in terms of \(U\) and \(t\), for
   1. \(x\),
   2. \(y\). [3]
2. The ball takes \(T\) s to travel from O to P. Show that \(T = \frac{U\sin 68.5°}{4.9}\) and write down a second equation connecting \(U\) and \(T\). [4]
3. Hence show that \(U = 12.0\) (correct to three significant figures). [3]
4. Calculate the horizontal distance of the ball from the platform when the ball lands on the ground. [5]
5. Use the expressions you found in part (i) to show that the cartesian equation of the trajectory of the ball in terms of \(U\) is $$y = x\tan 68.5° - \frac{4.9x^2}{U^2(\cos 68.5°)^2}.$$ Use this equation to show again that \(U = 12.0\) (correct to three significant figures). [4]

\includegraphics{figure_3} Fig. 7 shows the graph of \(y = \frac{1}{100}(100 + 15x - x^2)\). For \(0 \leq x < 20\), this graph shows the trajectory of a small stone projected from the point Q where \(y\) m is the height of the stone above horizontal ground and \(x\) m is the horizontal displacement of the stone from O. The stone hits the ground at the point R.

1. Write down the height of Q above the ground. [1]
2. Find the horizontal distance from O of the highest point of the trajectory and show that this point is \(1.5625\) m above the ground. [5]
3. Show that the time taken for the stone to fall from its highest point to the ground is \(0.565\) seconds, correct to 3 significant figures. [3]
4. Show that the horizontal component of the velocity of the stone is \(22.1\text{ms}^{-1}\), correct to 3 significant figures. Deduce the time of flight from Q to R. [5]
5. Calculate the speed at which the stone hits the ground. [4]

Sandy is throwing a stone at a plum tree. The stone is thrown from a point O at a speed of \(35\text{ms}^{-1}\) at an angle of \(\alpha\) to the horizontal, where \(\cos\alpha = 0.96\). You are given that, \(t\) seconds after being thrown, the stone is \((9.8t - 4.9t^2)\) m higher than O. When descending, the stone hits a plum which is \(3.675\) m higher than O. Air resistance should be neglected. Calculate the horizontal distance of the plum from O. [6]

In a fairground game, a contestant bowls a ball at a coconut 6 metres away on the same horizontal level. The ball is thrown with an initial speed of 8 ms\(^{-1}\) in a direction making an angle of 30° with the horizontal. \includegraphics{figure_8}

1. Find the time taken by the ball to travel 6 m horizontally. [2 marks]
2. Showing your method clearly, decide whether or not the ball will hit the coconut. [4 marks]
3. Find the greatest height reached by the ball above the level from which it was thrown. [4 marks]
4. Find the maximum horizontal distance from which it is possible to hit the coconut if the ball is thrown with the same initial speed of 8 m s\(^{-1}\). [3 marks]
5. State two assumptions that you have made about the ball and the forces which act on it as it travels towards the coconut. [2 marks]

A stone, of mass 1.5 kg, is projected horizontally with speed 4 ms\(^{-1}\) from a height of 7 m above horizontal ground.

1. Show that the stone travels about 4.78 m horizontally before it hits the ground. [4 marks]
2. Find the height of the stone above the ground when it has travelled half of this horizontal distance. [4 marks]
3. Calculate the potential energy lost by the stone as it moves from its point of projection to the ground. [2 marks]
4. Showing your method clearly, use your answer to part (c) to find the speed with which the stone hits the ground. [3 marks]
5. State two modelling assumptions that you have made in answering this question. [2 marks]