# Physics

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A basketball player is running at 5.00 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 0.750 meters above the floor?

(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?

• Physics -

a. Y^2 = Yo^2 + 2g*h = 0 m/s @ max. ht.
Yo^2 - 19.6*0.750 = 0
Yo^2 = 14.7
Yo = 3.83 m/s. = Initial ver. velocity.

b. Y = Yo + g*Tr = 0 m/s @ max. ht.
Y = 3.83 - 9.8Tr = 0
9.8Tr = 3.83
Tr = 0.3908 s. = Rise time.

Dx=Xo * Tr = 5m/s * 0.3908s. = 1.95 m.

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