college physics question?

Question

A 2.05 meter tall basketball player takes a shot when he is 6.02m from the basket (at the 3 point line). If the launch angle is 25 degrees and the ball was launched at the level of the players head, what much be the release speed of the ball for the player to make the shot? the basket is 3.05m above the floor.

Answer 1

The first step, is to consider the height of the player's head as the horizontal plane w/r/t dimensions. Using this frame of reference, the ball starts at y=0, x=0, and must fly to y=1, x=6.02
For all calculations I will ignore friction.

First, consider motion in the x direction:

x(t)=cos(25)*v*t
where v is the magnitude of the velocity of the ball

when x(t)=6.02
v=6.02/(cos(25)*t)

Motion in the y direction
y(t)=sin(25)*v*t-.5*g*t^2

When y(t)=1

0=.5*g*t^2-sin(25)*v*t+1

from the horizontal motion above

v=6.02/(cos(25)*t)

substitute into the y(t)=1

=-0.5*g*t^2+
SIN(25)*6.02/COS(25)-1=0

doing the algebra

t=.607 s
v=10.94 m/s

checking

x(.607)=6.02
y(.607)=1
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