Question

A basketball player crouches while waiting to jump for the ball. He lowers his center of gravity is 0.30 m. When he jumps for the ball, his center of gravity reaches a height of 0.90 m above its normal position.(a) Calculate the velocity of the player when his feet left the ground. (b) Calculate the acceleration he produced to achieve this velocity. (c) What force did he exert on the floor if his mass is 110 kg?

Answer 1

(a)
(1/2) m v^2 = m g h = m (9.8)(.9)
v^2 =2*9.8*.9
v = 4.2 m/s

(c)
work done to get from -.3 m to 0 meters where feet leave ground = F*.3 = change in potential + change in kinetic

F * .3 = m g (.3) + .5 m (4.2)^2
.3 F = 110 (.3*9.8 + .5*17.64)
F = 4312 N

(b)
F - mg = m a
4312 - 1078 = 110 * a
a = 29.4 m/s^2

wow, 3 g !

Answer 2

alternate (b)
we have v = 4.2
so average v during acceleration = 2.1
so time to go up .3 m = .3/2.1 = .143
change in v/time = a = 4.2/.143 = 29.4 again

Answer 3

Thank you Damon, It's appreciated