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A basketball player is running at 4.80 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 (in m/s) does he need to rise 0.650 meters above the floor? _______m/s (b) How far (in m) 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

1 Answer

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Answer:

a) 3.56m/s

b) 1.73m

Step-by-step explanation:

We have to treat this as a parabolic motion problem:

we will use the next formulas:


V=Vo+a*t\\Vy^2=Vyo^2+2*a*Y\\X=Vox*t

we first have to calculate the initial velocity of the basketball player:


Vy^2=Vyo^2+2*a*Y\\0^2=Vyo^2+2*(-9.8)*(0.650)\\Vyo=√(2*9.8*0.650) \\Vyo=3.56 m/s

the final velocity is zero when he reaches the maximun height.

To answer the second part we need to obtain the time to reach the maximun height, so:


V=Vo+a*t\\\\0=3.56+(-9.8)*t\\t=0.36 seconds

Now having that time, let's find the distance on the X axis, the X axis behaves as constant velocity movement, so:


X=Vox*t\\X=4.80*0.36\\X=1.73m

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