A ball launched at v0 = 20\text{ m/s} and \theta = 50° on level ground takes how long to return to launch height? Use g = 9.8\text{ m/s}^2.

3.13 s Time of flight: T = 2 v0 \sin\theta / g = 2(20)(\sin 50°)/9.8 \approx 2(20)(0.766)/9.8 \approx 3.13\text{ s}.

Vectors and Motion in Two Dimensions

AP Physics 1· difficulty 4/5

v_0 = 20\text{ m/s}

Explanation

T = 2 v_0 \sin\theta / g = 2(20)(\sin 50°)/9.8 \approx 2(20)(0.766)/9.8 \approx 3.13\text{ s}

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