Question

in a basketball game a referee throws the basketball upwards from an initial height of 2.00 meters above the floor, with sufficient initial speed such that its highest point will be 4.20 meters above the floor. a basketball players outstretched hand is 2.50 meters above the floor when he holds his hand above his head, and he can jump with an initial speed of 3.90m/s upwards. how long after the ball is released from the referees hand should the player jump in order to reachthe ball just as his highest point of his jump, if he is to do so (a) while the ball is still on its way up, and (b) when the ball is on its way back down?

Answer 1

When the refree throws the ball

intial height = 2 m

max. height = 4.2 m

max. raise h = 4.2 - 2.0 = 2.2 m

initial vel u = sqrt(2gh) = sqrt(2*9.8*2.2) = 6.57 m/s

vertical position of the ball at any instant of time

y_r(t) = 2+ ut -gt² /2 = 2+ 6.57t -9.8*t² /2

         = 2+6.57t - 4.9t²

initial height of the player = 2.5 m

intial vel = 3.9 m/s

The player will reach max height when his final vel =0

0 = 3.9 - 9.8t

t = 0.398 s

Player height at any insatnt of time t is

y_p(t) = 2.5 +3.9t - 4.9t²

For the player to catch the ball at some instant of time

y_r(t) = y_p(t)

2 + 6.57t - 4.9t² =  2.5 +3.9t - 4.9t²

t = 0.187 s

y_p(0.187) = y_r(0.187) = 3.057 m

The player shall jump immediately as the referee throws the ball so that he catches the ball after 0.187 s while it is on its way up.

for the ball to return to the same height while descending

2.5 +3.9t - 4.9t² = 3.057

t = 0.187, 0.609 s

For catching the ball while descending the player haas to jump after 0.609 - 0.187 = 0.422s