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IB Mathematics: Applications and Interpretation HL · AHL 4: Vectors · AHL 4.25

Projectiles: launch angle, launch height and clearing a wall Problem set B One question at a time

Try each question on paper first. Open the working a line at a time, then say whether you had it.

Skills

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.

In context

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.

Stretch

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.

jmaths.xyz · AI HL · AHL 4.25 Projectiles: launch angle, launch height and clearing a wall