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A basketball player is running at 5.20 m/s directly toward the basket when he jumps into the air to dunk the ball. He maintains his horizontal velocity.

(a) What vertical velocity does he need to rise 0.600 meters above the floor?

m/s

(b) How far from the basket (measured in the horizontal direction) must he start his jump to reach his maximum height at the same time as he reaches the basket?

m

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Added Thu, 28 Apr '16
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a) If you are talking about finding the initial vertical velocity (Vo), Then we can solve for it using this equation:

V^2 = Vo^2 - (2 x g x y) Solve for Vo

V = Final velocity

g = acceleration due to gravity

y = vertical distance traveled

V^2 = zero because V = 0 So:

Vo = square root (2 x g x y)

Vo = square root (2 x 9.80 x .60m) = 3.4292m/s

b) Before we can solve for the starting distance from the basket, we need to know the total time elapsed (t). The time it takes the player to reach maximum height is equal to the time it takes the player to travel horizontally from the take off point to the basket. Solve for (t) using this equation:

V = Vo - (g x t) Solve for t:

V = Final velocity = 0

Vo = Initial vertical velocity

g = acceleration due to gravity

t = time

t = Vo / g

t = 3.4292 / 9.80 = .349927sec

Since the horizontal velocity is constant throughout, we can multiply the horizontal velocity (Vh) by time (t) to find the distance (d)

d = Vh x t

d = 5.20m/s x .349927s = 1.8196m

Edit
Edited Thu, 28 Apr '16
Added Thu, 28 Apr '16
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