# A soccer ball leaves a cliff 20.2 m above the valley floor, at an angle of 10 degrees above the horizontal. The ball hits the valley floor 3.0 sec later. What is the initial velocity of the ball? What maximum height above the cliff did the ball reach?

Aug 31, 2017

$\textsf{\left(a\right) . 46.0 \textcolor{w h i t e}{x} \text{m/s}}$

$\textsf{\left(b\right) . 3.24 \textcolor{w h i t e}{x} m}$

#### Explanation:

Considering the vertical component of the motion we can use:

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

Using the convention "up is +ve" and setting the ball at the origin this becomes:

$\textsf{h = v \sin \theta t - \frac{1}{2} \text{g} {t}^{2}}$

Putting in the numbers:

$\textsf{- 20.2 = \left(v \sin 10 \times 3.0\right) - \frac{1}{2} \times 9.81 \times {3.0}^{2}}$

$\textsf{- 20.2 = \left(v \times 0.1736 \times 3.0\right) - 44.145}$

$\textsf{v = \frac{23.945}{0.5208} = \textcolor{red}{46.0 \textcolor{w h i t e}{x} \text{m/s}}}$

To find the height above the cliff we can use:

$\textsf{{v}^{2} = {u}^{2} + 2 a s}$

This becomes:

$\textsf{0 = {\left(v \sin \theta\right)}^{2} - 2 g s}$

$\textsf{0 = {\left(45.976 \times 0.1736\right)}^{2} - 2 \times 9.81 \times s}$

sf(s=63.487/19.62=color(red)(3.24color(white)(x)m)