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

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SUMMARY

The release speed of a basketball shot can be calculated using the motion equations for projectile motion. In this discussion, the player is 6.02 meters from the basket, which is 2.05 meters high, and the launch angle is 25 degrees. The correct initial velocity (V0) calculated is 8.78 m/s, assuming gravitational acceleration (g) is 9.8 m/s². The equations used are x = V0 cos(theta) t and y = V0 sin(theta) t - 1/2 g t², with y set to 0 m for the final position.

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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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