# A ball is thrown horizontally from a cliff at a speed of 15 m/s and strikes the ground 45 meters from the base of the cliff. How high was the cliff?

Nov 19, 2017

Assuming air resistance is negligible, we can state that the ball flew for 3 seconds ($t = \frac{d}{v}$)

#### Explanation:

This then allows us to use a kinematic equation (SUVAT equation) to solve for the vertical distance travelled. Initial (vertical) velocity, called ‘u’ below is zero as the ball is thrown horizontally.

We’ll use $s = u . t + \frac{1}{2} a . {t}^{2}$ knowing the acceleration to be 9.81 $\frac{m}{s} ^ 2$ (I guess the ball hits the earth, not another body!) and that u in the vertical direction is zero, we get $s = \frac{1}{2} a . {t}^{2}$

$s = \frac{1}{2} \times 9.81 \left({3}^{2}\right)$

$s = 44.145$ m

I’d quote this to 2 sig. figs., as the data in the question implies.