Question

A basketball player grabbing a rebound jumps 76.0 cm vertically. How much total time (ascent and descent) does the player spend...
(a) in the top 15.0 cm of this jump?
(b) in the bottom 15 cm? (The player seems to hang in the air at the top)
Answer both parts (a) and (b). Show all work and give explanation for calculations

Answer 1

Answer: Part(a)=0.041 secs, Part(b)=0.041 secs

Explanation: Firstly we assume that only the gravitational acceleration is acting on the basket ball player i.e. there is no air friction

now we know that

a=-9.81 m/s^2  ( negative because it is pulling the player downwards)

we also know that

s=76 cm= 0.76 m ( maximum s)

using kinetic equation

$v^2=u^2+2as$

where v is final velocity which is zero at max height and u is it initial

hence

$u^2=-2(-9.81)*0.76$

$u=3.8615 m/s$

now we can find time in the 15 cm ascent

$s=ut+0.5at^2$

$0.15=3.861*t+0.5*9.81t^2$

using quadratic formula

$t=\frac{-3.861+\sqrt{3.86^2-4*0.5*9.81(-0.15)} }{2*0.5*9.81}$

t=0.0409 sec

the answer for the part b will be the same

To find the answer for the part b we can find the velocity at 15 cm height similarly using

$v^2=u^2+2as$

where s=0.76-0.15

as the player has traveled the above distance to reach 15cm to the bottom

$v^2=0^2 +2*(9.81)*(0.76-0.15)$

$v=3.4595$

when the player reaches the bottom it has the same velocity with which it started which is 3.861

hence the time required to reach the bottom 15cm is

$t=\frac{3.861-3.4595}{9.81}$

t=0.0409