A basketball player jumps straight up for a ball. To do this, he lowers his body 0.310 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 0.920 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 (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 110.0 kg.

a)  v₀ = 4.25 m / s , b)  a = 30.1 m / s², c) F = 3311 N

Explanation:

a) to calculate the speed with which it leaves the ground we use the kinematic relations

        v² = v₀² - 2 g y

where the speed at the highest point is zero (v = 0) and the height is y = 0.920m, this implies that our reference system is on the ground

        v₀ = √ 2gy

let's calculate

       v₀ = √(2 9.8 0.920)

       v₀ = 4.25 m / s

b) to find the acceleration to reach the speed of v = 4.25 m over a distance of y = 0.300 m

       v² = v₀² + 2 a y

in this case it starts from an initial velocity of zero

       v² = 2 a y

       a = v² / 2y

let's calculate

       a = 4.25² / (2 0.300)

       a = 30.1 m / s²

c) to calculate the force we use Newton's second law

        F = m a

     let's calculate

        F = 110.0 30.1

        F = 3311 N

for this one it would be d bump/forearm pass

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