Calculate Horizontal Ball Drop at 41.3m/s Velocity

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SUMMARY

The discussion focuses on calculating the vertical drop of a baseball thrown horizontally at a velocity of 41.3 m/s over a distance of 17.0 meters. The key equations involved include the vertical motion equation, where Ay is -9.8 m/s², and the horizontal motion equation, where the initial horizontal speed (Vox) is 41.3 m/s. The time (t) it takes for the ball to reach the catcher can be determined using the formula t = X / Vox, resulting in a time of approximately 0.412 seconds. The vertical drop can then be calculated using the formula y = (1/2) * Ay * t².

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pookisantoki
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A Major league pitcher can throw a ball in excess of 41.3m/s. If a ball is thrown horizontally at this speed, how much will it drop by the time it reaches a catcher who is 17.0m away from the point of release?
So Ay=-9.8
X=17m
t=?
Vox=0
V=41.3
y=(Vy^2-Voy^2)2g
I thoguht i would use this formula but wasnt sure how to figure Voy out since Vy=0...
Please help!
 
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pookisantoki said:
X=17m
t=?
Vox=0

Vox, the initial horizontal speed, is not 0, it is ____?
Also, ax = ____?

Can you fill in the blanks, and then use this information to find t?
 

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