Projectiles: launch angle, launch height and clearing a wall Problem set B Worked solutions

A marble is thrown horizontally at 1.5 m s⁻¹ from a height of 1.2 m above level ground.

1. (a) Find the time it takes to reach the ground.

   • Vertical position \(y = 1.2 − 4.9t^{2}\) (it starts with no vertical speed).
   • Solve \(1.2 − 4.9t^{2} = 0\): \(t = 0.495\) s.

2. (b) Find the horizontal distance from the launch point to where it lands.

   • Horizontal position \(x = 1.5t\).
   • \(x = 1.5 \times 0.4949 = 0.742\) m.

A football is projected from the origin with speed 30 m s⁻¹ at 35° above the horizontal. Take \(\mathbf{i}\) as horizontal and \(\mathbf{j}\) as vertically up.

1. (a) Write down the horizontal and the vertical component of the initial velocity.

   • \(30\cos 35^\circ = 24.6\), \(30\sin 35^\circ = 17.2\).

2. (b) Write down the velocity vector at time \(t\) seconds.

   • Horizontal: constant. Vertical: decreases by 9.8 every second.
   • \(\mathbf{v} = \begin{pmatrix} 24.57 \\ 17.21 − 9.8t \end{pmatrix}\).

3. (c) Write down the position vector at time \(t\) seconds.

   • Integrate, with zero constants because it starts at the origin.
   • \(\mathbf{r} = \begin{pmatrix} 24.57t \\ 17.21t − 4.9t^{2} \end{pmatrix}\).

A basketball is projected from level ground with speed 11 m s⁻¹ at 60° above the horizontal.

1. (a) Find the time it takes to reach its highest point.

   • Vertical component \(11\sin 60^\circ = 9.526\).
   • \(9.526 − 9.8t = 0\), \(t = 0.972\) s.

2. (b) Find its greatest height.

   • \(y = 9.526(0.9721) − 4.9(0.9721)^{2} = 4.63\) m.

3. (c) Find the horizontal distance from the launch point to where it lands.

   • Landing time \(2 \times 0.9721 = 1.944\) s.
   • Horizontal component \(11\cos 60^\circ = 5.5\).
   • \(x = 5.5 \times 1.944 = 10.7\) m.

A ball is projected from a point 5 m above level ground with speed 12 m s⁻¹ at 35° above the horizontal.

1. (a) Find the time it takes to reach the ground.

   • Vertical position \(y = 5 + 6.883t − 4.9t^{2}\).
   • Polynomial solver: \(y = 0\), \(t = 1.93\) s (the positive solution).

2. (b) Find the horizontal distance from the launch point to where it lands.

   • Horizontal component \(9.830\).
   • \(x = 9.830 \times 1.933 = 19.0\) m.

3. (c) Find its greatest height above the ground.

   • Top at \(6.883 − 9.8t = 0\), \(t = 0.7023\) s.
   • \(y = 5 + 6.883(0.7023) − 4.9(0.7023)^{2} = 7.42\) m.

A pebble is thrown from a pier. It is thrown horizontally at 9 m s⁻¹ and hits the water 1.8 seconds later. Take \(\mathbf{i}\) as horizontal and \(\mathbf{j}\) as vertically up.

1. (a) Find its velocity as a column vector when it hits the water.

   • \(\mathbf{v} = \begin{pmatrix} 9 \\ −9.8t \end{pmatrix}\), so \(\mathbf{v}(1.8) = \begin{pmatrix} 9 \\ −17.64 \end{pmatrix}\).

2. (b) Find its speed when it hits the water.

   • \(|\mathbf{v}| = \sqrt{9^{2} + 17.64^{2}} = 19.8\) m s⁻¹.

3. (c) Find the angle below the horizontal at which it hits the water.

   • \(\tan\theta = \dfrac{17.64}{9}\), \(\theta = 63.0\)° below the horizontal.

A football is kicked from level ground with speed 18 m s⁻¹ at 40° above the horizontal, towards a net 5.2 m high that is 22 m away along the ground. Take \(\mathbf{i}\) as horizontal and \(\mathbf{j}\) as vertically up, with the origin at the launch point.

1. (a) Find the time at which the ball is above the net.

   • Horizontal component \(13.79\).
   • \(13.79t = 22\), \(t = 1.595\) s.

2. (b) Find the height of the ball at that time.

   • \(y = 11.57(1.595) − 4.9(1.595)^{2} = 5.99\) m.

3. (c) Does the ball clear the net? Give the difference between its height and the top of the net.

   • \(5.99 > 5.2\).
   • The ball is 0.787 m above the top: it clears the net.

A skier leaves horizontally at 11 m s⁻¹ from a point 8 m above level water.

1. (a) Find the time it takes to reach the water.

   • \(8 − 4.9t^{2} = 0\), \(t = 1.28\) s.

2. (b) Find how far from the point directly below the launch it lands.

   • \(x = 11 \times 1.278 = 14.1\) m.

3. (c) Find its speed and the angle below the horizontal as it reaches the water.

   • \(\mathbf{v} = \begin{pmatrix} 11 \\ −9.8t \end{pmatrix}\), \(\mathbf{v}(1.278) = \begin{pmatrix} 11 \\ −12.5 \end{pmatrix}\).
   • \(|\mathbf{v}| = 16.7\) m s⁻¹, \(\tan\theta = \dfrac{12.5}{11}\), \(\theta = 48.7\)°.

A shot put is released from a point 2.1 m above level ground with speed 13 m s⁻¹ at 38° above the horizontal. Take \(\mathbf{i}\) as horizontal and \(\mathbf{j}\) as vertically up, with the origin on the ground below the release point.

1. (a) Write down the position vector at time \(t\) seconds.

   • \(13\cos 38^\circ = 10.24\), \(13\sin 38^\circ = 8.004\).
   • \(\mathbf{r} = \begin{pmatrix} 10.24t \\ 2.1 + 8.004t − 4.9t^{2} \end{pmatrix}\).

2. (b) Find its greatest height above the ground.

   • \(8.004 − 9.8t = 0\), \(t = 0.8167\) s.
   • \(y = 2.1 + 8.004(0.8167) − 4.9(0.8167)^{2} = 5.37\) m.

3. (c) Find the time it takes to land and the horizontal distance of the landing point from the origin.

   • Polynomial solver: \(y = 0\), \(t = 1.86\) s.
   • \(x = 10.24 \times 1.863 = 19.1\) m.

A school team practises kicking over a fence. The ball is kicked from level ground with speed 17 m s⁻¹ at 32° above the horizontal. The fence is 2.4 m high and stands 12 m from the kick. Take \(\mathbf{i}\) as horizontal and \(\mathbf{j}\) as vertically up, with the origin at the kick.

1. (a) Find the height of the ball when it is 12 m from the kick, and say whether it clears the 2.4 m fence.

   • Time at 12 m: \(14.42t = 12\), \(t = 0.8324\) s.
   • \(y = 9.009(0.8324) − 4.9(0.8324)^{2} = 4.10\) m, which is above 2.4 m: it clears the fence.

2. (b) Find the speed of the ball at that moment.

   • \(\mathbf{v} = \begin{pmatrix} 14.42 \\ 9.009 − 9.8t \end{pmatrix}\), \(\mathbf{v}(0.8324) = \begin{pmatrix} 14.42 \\ 0.851 \end{pmatrix}\).
   • \(|\mathbf{v}| = 14.4\) m s⁻¹.

3. (c) Find how far from the kick the ball lands.

   • Landing time \(2 \times 0.9192 = 1.838\) s.
   • \(x = 14.42 \times 1.838 = 26.5\) m.

A golf ball is projected from level ground with speed 35 m s⁻¹ at 40° above the horizontal.

1. (a) Find the two times at which the ball is 10 m above the ground.

   • \(y = 22.50t − 4.9t^{2}\).
   • Polynomial solver: \(22.50t − 4.9t^{2} = 10\), \(t = 0.499\) or \(t = 4.09\). One time is on the way up, one on the way down.

2. (b) Find the horizontal distance from the launch point at each of these times.

   • \(x = 26.81t\): \(26.81 \times 0.4986 = 13.4\) m and \(26.81 \times 4.093 = 110\) m.

3. (c) For how long is the ball more than 10 m above the ground?

   • It is above 10 m between the two times: \(4.093 − 0.4986 = 3.59\) s.

A ball is kicked from level ground at an angle of 45° above the horizontal, with speed \(V\) m s⁻¹, towards a wall 6 m high that is 30 m away along the ground. Take \(\mathbf{i}\) as horizontal and \(\mathbf{j}\) as vertically up, with the origin at the kick.

1. (a) Show that the height of the ball when it is 30 m from the kick is \(30 − \dfrac{8820}{V^{2}}\) metres.

   • \(\mathbf{r} = \begin{pmatrix} \dfrac{V}{\sqrt{2}}t \\ \dfrac{V}{\sqrt{2}}t − 4.9t^{2} \end{pmatrix}\).
   • \(\dfrac{V}{\sqrt{2}}t = 30\), so \(t = \dfrac{30\sqrt{2}}{V}\).
   • \(y = 30 − 4.9\left(\dfrac{30\sqrt{2}}{V}\right)^{2} = 30 − \dfrac{8820}{V^{2}}\).

2. (b) Find the least value of \(V\) for which the ball clears the wall.

   • The ball just clears the wall when \(30 − \dfrac{8820}{V^{2}} = 6\).
   • Solve on the calculator: \(V = 19.2\) m s⁻¹ (positive solution).
   • A larger \(V\) gives a greater height, so this is the least value.