Projectile Motion - baseball popup

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SUMMARY

The discussion focuses on calculating the launch angle of a baseball that remains airborne for 6.3 seconds and lands 83 meters away. The participant derived vertical velocity (Vy) as 30.87 m/s and horizontal velocity (Vx) as 13.17 m/s. Using the relationship between Vy and Vx, the launch angle (Q) was determined to be 66 degrees. The calculations utilized the equations for projectile motion, specifically Vx = V * cos(Q) and Vy = V * sin(Q).

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Projectile Motion -- baseball popup

Homework Statement


A baseball is popped up, remaining aloft for 6.3 s before being caught at a horizontal distance of 83 m from the starting point. What was the launch angle?


Homework Equations


vx = vcostheta
vy = vcostheta

The Attempt at a Solution


I tried solving for vx and vy then take those and find the arctan but then there was an erroron my caclculator. Besides, I think I'm doing something wrong.
 
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Vx= V*CosQ
Vy=V*SinQ
t= 2*Vy/g=6.3 =>Vy=30.87 m/s
Range= Vx*t=83m => Vx=13.17 m/s
Vy/Vx=TanQ=66 Degrees
 

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