A basketball player standing near the basket to grap a rebound, jumps 62 cm vertically. How much total time does the player s in the top 15 cm of the jump? How much total time does the player spend in the bottm 15 cm of this jump?

Answer 1

If he jumps up H = 0.62 m, the initial velocity V is sucn that
(1/2) M V^2 = M g H
V = sqrt (2gH) = 3.48 m/s

Height vs time is given by
y = 3.38 t - 4.90 t^2
Solve for the times when H = 0.15 m and y = 0.47 m. You will need to use the postive root of the quadratic eauation.

At the top of the trajectory, H = 0.62 m and g t = V = 3.48 m/s, so t = 0.355 s there.
t = 0 when y = 0. Knowing the times when H = 0.15 and 0.47 m will let you calculate the intervals they are asking for.