Question

A basketball player can jump vertically 0.9 m. assume that the jump interval (from the lowering the body to his feet taking off) lasts for 0.3 s and the mass of the player is 90 kg. ignore air resistance. determine the force the player exerts on the surface when he jumps. express your answer to the significant figure and include the appropriate units. enter positive value if the force is upward and negative value if the force is downward. determine the force that the surface exerts on the player when he jumps. express your answer to one significant figure and include the apropriate units. enter positive value if the force is upward and negative value if the force is downward.determine the force that the surface exerts on the player when he is landing back on the floor. express your answer to one significant figure and include the appropriate units. enter positive value if the force is upward and negative value if the force is downward. how would the answers change if the air resistance is considered?

Answer 1

Impulse: I = Ft = ∆p = m∆v

To calculate the impulse one needs to find the change in momentum.
The speed of the player as he leaves the ground: Vf² = Vi² - 2gh =>
Vi² = Vf² + 2gh
Vi = √(2gh) = √(2 * 9,8m/s² * 0.9m) = 4.2 m/s = ∆v

Momentum ∆p = m∆v = 90kg * 4.2m/s = 378 kgm/s
F = ∆p/t = 378kgm/s / 0.3s = 1260 N

This is the impulse needed to give the player the required speed
to reach 0.9 m. But he must also lift his own weight of the ground
as well.
F = mg = 90kg * 9.8m/s² = 882 N

Total force: 1260N + 882N = 2142 N
(The negative sign is due to the direction. Probably up is the positive direction
and the player is pushing with a downward force on the ground.)