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Question

A basketball player grabbing a rebound jumps 76.0  cm76.0\text{~ cm} vertically. How much total time (ascent and descent) does the player spend in the top 15.0 cm15.0 \text{~cm} of this jump ? (The player seems to hang in the air at the top.)

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Answered 1 year ago
Answered 1 year ago
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Since there is constant acceleration you can use the following equation to find the time of the top 15 cm of the player's descent:

xfxi=vit+12at20.61.76=012(9.8)t2t.175s\begin{align*}x_{f} - x_{i}& = v_{i}t + \dfrac{1}{2} a t^{2}\\0.61 - .76& = 0 - \dfrac{1}{2} (-9.8) \cdot t^{2}\\t& \approx .175 \: \rm{s}\end{align*}

Multiply this answer by 2 to find the total time for ascending and descending in the top 15 cm.

Total of top 15 cm ascent + descent .35s\approx .35 \: \rm{s}

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