How Much Does a Baseball Drop When Thrown Horizontally at 46.8 m/s Over 15.7 m?

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SUMMARY

The discussion focuses on calculating the vertical drop of a baseball thrown horizontally at a speed of 46.8 m/s over a distance of 15.7 m. The time taken for the ball to reach the catcher is determined to be 0.33 seconds using the formula v = d/t. The correct formula for vertical displacement, accounting for gravitational acceleration, is applied to find that the baseball drops approximately 15.98 meters by the time it reaches the catcher.

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Homework Statement



A major-league pitcher can throw a baseball in excess of 46.8 m/s. If a ball is thrown horizontally at this speed, how much will it drop by the time it reaches the catcher who is 15.7 m away from the point of release?


Homework Equations





The Attempt at a Solution



i first found time using v=d/t
so 46.8 = 15.7/ t
t= 0.33

then i used the formula x=x_0 +v_0*t
so x=15.7+46.8*0.33
x= 31.14
 
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You have the time, but you use the wrong distance formula. Remember that gravity is constantly accelerating the ball downwards.
 
ok y=y_0 + v_0*t-g/2*t^2
y=0+46.8*0.33-g/2*t^2
y=15.98
 

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