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# 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?

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#### Explanation

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#### Explanation:

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Mark C. Share
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.

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