Vectors

Have a go first. Then open the working one line at a time.

1. Position

   \(\mathbf{r} = \begin{pmatrix} 15t \\ 30 − 4.9t^{2} \end{pmatrix}\)

   Horizontally the speed is constant. Vertically it starts with speed 0 and falls under gravity, starting at height 30.

2. Time to land

   \(30 − 4.9t^{2} = 0\), \(t = 2.47\) s

   On the ground the height is zero. Take the positive solution.

3. Distance

   \(x = 15 \times 2.474 = 37.1\) m

   Horizontal distance = horizontal speed × time.

4. Velocity on landing

   \(\mathbf{v} = \begin{pmatrix} 15 \\ −9.8t \end{pmatrix}\), so \(\mathbf{v} = \begin{pmatrix} 15 \\ −24.2 \end{pmatrix}\) m s⁻¹

   Differentiate the position. Put the landing time into the vertical component.

5. Speed and direction

   \(|\mathbf{v}| = \sqrt{15^{2} + 24.2^{2}} = 28.5\) m s⁻¹, \(\tan\theta = \dfrac{24.2}{15}\), \(\theta = 58.3\)° below the horizontal

   The vertical component is negative, so the stone is moving downwards.

Have a go first. Then open the working one line at a time.

1. Components

   \(\mathbf{v}(0) = \begin{pmatrix} 22\cos 35^\circ \\ 22\sin 35^\circ \end{pmatrix} = \begin{pmatrix} 18.02 \\ 12.62 \end{pmatrix}\)

   Horizontal part \(u\cos\theta\), vertical part \(u\sin\theta\). Calculator in degrees.

2. Velocity

   \(\mathbf{v} = \begin{pmatrix} 18.02 \\ 12.62 − 9.8t \end{pmatrix}\) m s⁻¹

   Gravity changes only the vertical component.

3. Top of the flight

   \(12.62 − 9.8t = 0\), \(t = 1.29\) s

   At the highest point the vertical velocity is zero.

4. Greatest height

   \(y = 12.62(1.288) − 4.9(1.288)^{2} = 8.12\) m

   Put the time of the top into the vertical displacement.

5. Range

   \(t_{land} = 2 \times 1.288 = 2.575\) s, \(x = 18.02 \times 2.575 = 46.4\) m

   On level ground the time up equals the time down. Horizontal distance = horizontal speed × total time.

Have a go first. Then open the working one line at a time.

1. Position

   \(\mathbf{r} = \begin{pmatrix} 10.72t \\ 2 + 8.999t − 4.9t^{2} \end{pmatrix}\)

   The vertical position starts at 2, not 0.

2. Time to land

   Polynomial solver: \(2 + 8.999t − 4.9t^{2} = 0\), \(t = 2.04\) s

   The other solution is negative: it is before the release.

3. Distance

   \(x = 10.72 \times 2.037 = 21.8\) m

   Horizontal speed × time in the air.

4. Greatest height

   \(8.999 − 9.8t = 0\), \(t = 0.918\), \(y = 2 + 8.999(0.9183) − 4.9(0.9183)^{2} = 6.13\) m

   Find the time of the top, then the height. Add the 2 m that it started with.

5. Landing velocity

   \(\mathbf{v}(2.037) = \begin{pmatrix} 10.72 \\ 8.999 − 9.8(2.037) \end{pmatrix} = \begin{pmatrix} 10.72 \\ −11.0 \end{pmatrix}\)

   Put the landing time into the velocity.

6. Speed and direction

   \(|\mathbf{v}| = 15.3\) m s⁻¹, \(\tan\theta = \dfrac{11.0}{10.72}\), \(\theta = 45.6\)° below the horizontal

   Its speed is greater than 14 m s⁻¹ because it lands lower than it started.

Have a go first. Then open the working one line at a time.

1. Position

   \(\mathbf{r} = \begin{pmatrix} 17.32t \\ 10t − 4.9t^{2} \end{pmatrix}\)

   Components: \(20\cos 30^\circ = 17.32\) and \(20\sin 30^\circ = 10\).

2. Time at the wall

   \(17.32t = 28\), \(t = 1.617\) s

   Use the horizontal component to find when the ball is 28 m away.

3. Height then

   \(y = 10(1.617) − 4.9(1.617)^{2} = 3.36\) m

   Put that time into the vertical displacement.

4. Compare

   \(3.36 > 3\)

   The ball is 0.360 m above the top of the wall, so it clears the wall.

5. Rising or falling

   \(v_{y} = 10 − 9.8(1.617) = −5.84\)

   The vertical velocity is negative, so the ball is on its way down as it goes over.