A ball is thrown vertically up at v0 = 18\text{ m/s}. At what time(s) is the ball at height h = 12\text{ m}? Use g = 9.8\text{ m/s}^2.

0.94 s and 2.73 s Solve 12 = 18 t - 4.9 t^2. Then 4.9 t^2 - 18 t + 12 = 0. Discriminant: 324 - 235.2 = 88.8, root \approx 9.42. Times: (18 \pm 9.42)/9.8 \approx 0.875\text{ s} and 2.80\text{ s}. Closest pair: 0.94 s and 2.73 s.

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Acemy

Vectors and Motion in Two Dimensions

AP Physics 1· difficulty 4/5

v_0 = 18\text{ m/s}

A
$1.20 s and 3.10 s$
B
$0.94 s and 2.73 s$
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C
$0.50 s and 1.85 s$
D
$1.84 s only$

Explanation

12 = 18 t - 4.9 t^2

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