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Archy [21]
3 years ago
13

a basketball player can leap upward .65m how long does the basketball player remain in the air use 9.81m/s²​

Physics
1 answer:
tigry1 [53]3 years ago
6 0

At the player's maximum height, their velocity is 0. Recall that

{v_f}^2-{v_i}^2=2a\Delta y

which tells us the player's initial velocity v_i is

0^2-{v_i}^2=-2g(0.65\,\mathrm m)\implies v_i=3.6\dfrac{\rm m}{\rm s}

The player's height at time t is given by

y=v_it-\dfrac g2t^2

so we find their airtime to be

0.65\,\mathrm m=\left(3.6\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2\implies t=0.36\,\mathrm s

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