Question

A basketball player grabbing a rebound jumps 64 cm vertically.

(a) How much (total) time does the player spend in the top 24 cm of this jump?
  ms

(b) How much (total) time does the player spend in the bottom 24 cm of this jump?
ms

Answer 1

Part A.

Using third kinematic equation:

V^2 = U^2 + 2*g*h

V = 0, since at the top velocity of player is zero, So initial velocity of player to reach this height (64 cm = 0.64 m) will be

U = sqrt (-2*g*h)

U = sqrt (-2*(-9.81)*0.64) = 3.54 m/sec

Now when player reaches the height of 24 cm (from top), his speed will be

At this time height from ground = 64 - 24 = 40 cm = 0.4 m

V1 = sqrt (U^2 + 2*g*h)

V1 = sqrt (3.54^2 + 2*(-9.81)*0.4) = 2.16 m/sec

Now Using equation

h = 0.5*(V1 + V)*t

t = 2h/(V1 + V)

t = 2*0.24/(2.16 + 0) = 0.222 sec

Now total time in the top of 24 cm = 2*t = 2*0.222 = 0.444 sec = 444 ms

Part B.

time spend in bottom of 24 cm will be given by:

Using 2nd kinematic equation

h = U*t1 + 0.5*g*t1^2

0.24 = 3.54*t1 - 0.5*9.81*t1^2

4.905*t1^2 - 3.54*t1 + 0.24 = 0

Solving above quadratic equation and taking smaller +ve root

t1 = [3.54 - sqrt (3.54^2 - 4*4.905*0.24)]/(2*4.905)

t1 = 0.0757 sec

So, total time spend in bottom of 24 cm = 2*0.0757 = 0.1514 sec = 151.4 ms

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