How do I find this 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

Yes, it does. Assuming no air resistance, the player has some initial velocity such that
x = vot -1/2 gt^2, where x=76 cm and g=980 cm/sec^2. You can solve the quadratic to find t, which is the time of the jump up. You can also write a quadratic for x=61 cm, and solve to find the time it takes for the player to reach this point. The difference is the time spent going from 61 to 76 cm. You also need to solve for 15 cm to do part of part (b)

Once he is at that point, he has no upward velocity left, and drops such that x= 1/2 gt^2. You again solve for 15 cm for the part (a) and then for 61 cm and for 76 cm. The time to 15 cm is added to the 61-76 cm portion of the upward leap. The time between 61 and 76 cm is added to the time for the initial 15 cm of the leap in part (a).