A ball is thrown horizontally from a cliff of height h = 45\text{ m} with speed v0 = 12\text{ m/s}. How far from the base does it land? Use g = 9.8\text{ m/s}^2.

36.4 m Time to fall: t = \sqrt{2h/g} = \sqrt{90/9.8} \approx 3.03\text{ s}. Horizontal distance: x = v0 t \approx 12 \times 3.03 \approx 36.4\text{ m}.

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Acemy

Vectors and Motion in Two Dimensions

AP Physics 1· difficulty 4/5

h = 45\text{ m}

A
$44.0 m$
B
$21.5 m$
C
$36.4 m$
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D
$29.7 m$

Explanation

t = \sqrt{2h/g} = \sqrt{90/9.8} \approx 3.03\text{ s}

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