Question

Integrated Concepts A basketball player jumps straight up for a ball. To do this, he lowers his body 0.300 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.900 m above the floor.
a. Calculate his velocity when he leaves the floor.
b. Calculate his acceleration while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.300 m.
c. Calculate the force he exerts on the floor to do this, given that his mass is 110 kg.

Answer 1

Answer:

a) Velocity = 4.2m/s

b) Acceleration = 2.94m/s^2

c) Force exerted on the floor= 1401.4×10^3N

Explanation:

a) Velocity,V=sqrt(2×9.8×0.900)

V= 4.2m/s

b) Vf2= V^2+2ay2

a= 4.2^2 - 0/2×3

a= 17.64/6= 2.94m/s^2

c) Newton's 2nd law indicates:

Fnet= F - mg=ma

F= m(g+a)

F=110(9.8+2.94)

F=110×12.94

F= 1401.4N