Physics question??????

Question

A basketball player, standing near the basket to grab a rebound, jumps 65.3 cm vertically. How much time does the player spend in the top 14.9 cm of his jump?

How much time does the player spend in the bottom 14.9 cm of the jump?

Answer 1

Consider the top 14.9 cm in the upward motion :

v = 0

s = 14.9 cm = 0.149 m

a = - 9.8

s = vt - at^2 / 2

Solve for t :

t = 0.17 s

So total time spent in the top 14.9 cm = 2 * t = 0.34 s

(Since the time to go up a distance of 14.9 cm is equal to the time taken to come down a distance of 14.9 cm)

Now consider the downward motion:

Distance traveled before covering the bottom 14.9 cm = 65.3 - 14.9 = 50.4 cm

So for the part of the motion before the bottom 14.9 cm:

s = 0.504 m

u = 0

a = 9.8

v^2 = u^2 + 2as

So v = 3.14

Now for the bottommost 14.9 cm :

s = 14.9 cm = 0.149 m

a = 9.8

u = 3.14

s = ut + 1/2*at^2

Solve for t.

Total time spent in the bottom 14.9 cm of the jump = 2t

Hope this helps.

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