How Do You Calculate the Release Speed of a Basketball Shot?

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To calculate the release speed of a basketball shot, the motion equations for horizontal and vertical components must be utilized. The player is 6.02 m from the basket, with the basket height at 2.05 m and a launch angle of 25 degrees. The calculated initial velocity (v0) is 8.78 m/s, assuming gravitational acceleration (g) is 9.8 m/s². The equations x = v0 cos(theta) t and y = v0 sin(theta) t - 1/2 g t² are essential for solving the problem. Properly applying these equations will yield the correct release speed.
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I am having a difficult time with the motion equation when the problem only includes an X value and an angle. The problem is to fin the velease speed of the ball when a player makes a shot to a basket when he is 6.02 m from the basket and the basket is 2.05 m above the floor. The player is 2.05 m tall and the launch angle is 25 degrees.

Attached is the method I used to do the calculation. I feel like I left something out.
 

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I can't see your work yet, but I got v0 = 8.78 m/s. Hope it helps. I'll check back later.

-Dan
 
V0=8.78m/s is right if we take g=9.8m/s^2.

by using
x=V0 cos(theta) t
y=V0 sin(theta) t - 1/2 g t^2

solving these two equations, you'll get the answer! remember y=0 m
 
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