Question

A basketball player, standing near the basket to grab a rebound, jumps 74.5cm vertically, how much time does the player spend in the top 15.3cm of his jump? How much time does he spend in the bottom 15.3cm of his jump?

Answer 1

Using equation

v^2 = u^2 + 2*a*s

v= u*t + 0.5*a*t^2

v = u + a*t

a = g

h = height = 74.5 cm = 0.745 m

g = 9.81 m/sec^2

v = at the top = 0

using first equation

0^2 = u^2 - 2*9.81*0.745

u = sqrt (2*9.81*0.745) = 3.823 m/sec

u = initial velocity

s = 15.3 cm = 0.153 m

similarly velocity to reach at the height of s

u1 = sqrt (2*9.81*0.153) = 1.732 m/sec

Now total time to reach at top

v = u + a*t

t = (v - u)/a

t = 2.823/9.81

t = 0.288 sec

total time of flight T = Tup + Tdown = 2*0.288 = 0.576 m

time to go from s to h

t = (v - u1)/a

t = 1.732/9.81 = 0.176 sec

Time spend above 15.3 cm will be

t1 = 2*0.176 = 0.352 sec

time spend below 15.3 cm

t2 = T - t1 = 0.576 - 0.352 = 0.224 sec