Figuring out a certain hang time...?

Question

A basketball player, standing near the basket to grab a rebound, jumps 76.0 cm vertically. How much total time does the player spend (a) in the top 15.0 cm of this jump and (b) in the bottom 15.0 cm? Does this help explain why such players seem to hang in the air at the tops of their jumps? Show and explain your work. Thanks and good luck! :)

Answer 1

s1=0.76 m, a = -9.81 m/s^2. The jump requires initial velocity v0 = sqrt(-2as1) = 3.862 m/s and reaches peak height s=0.76 in (v1-v0)/a = 0.3936 sec. The same v0 reaches s2 =s1-0.15 = 0.61 m, and velocity v1 = sqrt(v0^2+2as2) = 1.716 m/s in (v1-v0)/a = 0.2188 s.
ans. (a): Thus 2*(0.3936-0.2188) = 0.3496 s of the total 2*0.3936 = 0.7872 s flight time, or ~44%, is spent in the top 15/76 = ~20% of the jump, which does help explain "hanging".
ans. (b): There are two intervals in the jump spent in the bottom 15 cm. At s3=15 cm, each has a velocity v2 = sqrt(v0^2+2as3) = 3.46 m/s and spends t = (v2-v0)/a = 0.041 s in that region for a total 0.082 s.