# A ball is fired at a speed of 8 m/s at 20˚ above the horizontal from the top of a 5 m high wall. How far from the base of the ball will the ball hit the ground? Also calculate the angle below horizontal the velocity vector makes.

Feb 27, 2017

$\textsf{10 \textcolor{w h i t e}{x} m}$

$\textsf{{53.9}^{\circ}}$

#### Explanation: $\textsf{\left(a\right)}$

To get the time of flight we can use:

$\textsf{s = u t + \frac{1}{2} a {t}^{2}}$

$\textsf{{v}_{y} = 8 \sin 20}$

Using the convention that "down is +ve" this becomes:

$\textsf{5 = - 8 \sin 20 t + \frac{1}{2} \times 9.81 \times {t}^{2}}$

This simplifies to:

$\textsf{4.905 {t}^{2} - 2.736 t - 5 = 0}$

$\textsf{t = \frac{2.763 \pm \sqrt{7.634 - 4 \times 4.905 \times - 5}}{9.81}}$

Ignoring the -ve root this gives:

$\textsf{t = 1.33 \textcolor{w h i t e}{x} s}$

We can now get the range d from the horizontal component of velocity which is constant:

$\textsf{{v}_{x} = 8 \cos 20 = \frac{d}{1.33}}$

$\therefore$$\textsf{d = 8 \cos 20 \times 1.33}$

$\textsf{d = 7.517 \times 1.33}$

$\textsf{d = 10 \textcolor{w h i t e}{x} m}$

$\textsf{\left(b\right)}$

To get the angle to the horizontal which the ball strikes the ground we need the vertical component of its velocity when it lands.

We can use:

$\textsf{v = u + a t}$

This becomes:

$\textsf{{v}_{y} = - 8 \sin 20 + \left(9.81 \times 1.33\right)}$

$\textsf{{v}_{y} = - 2.736 + 13.034 = 10.3 \textcolor{w h i t e}{x} \text{m/s}}$

See fig (b):

We now know:

$\textsf{{v}_{x} = 7.517 \textcolor{w h i t e}{x} \text{m/s}}$

$\textsf{{v}_{y} = 10.3 \textcolor{w h i t e}{x} \text{m/s}}$

If we needed the resultant we could use Pythagoras, however we are only asked for the angle to the horizontal $\alpha$.

If $\textsf{\theta}$ is the angle to the vertical then:

$\textsf{\tan \theta = \frac{7.517}{10.3} = 0.7298}$

$\therefore$$\textsf{\theta = {\text{tan}}^{- 1} 0.7298 = {36.12}^{\circ}}$

So the angle to the horizontal is given by:

$\textsf{\alpha = {90}^{\circ} - {36.12}^{\circ} = {53.87}^{\circ}}$