A major-league pitcher can throw a ball in excess of 41.4 m/s. If a ball is thrown horizontally at this speed, how much will it drop by the time it reaches a catcher who is 17.0 m away from the point of release?

Question

anonymous · Answer

First you need to find the time it takes to reach the pitcher. By using the formula d=rt, the time is 17/41.4 or $$\approx.411 seconds$$ This is the same time you can use to find out how far it fell. Since horizontal and vertical velocities are separate and don't affect each other, you can ignore the throw speed for now and focus just focus on gravity, which is pulling the ball straight down. The formula for distance due to acceleration is $$d=.5a t^{2}$$ by plugging in 9.8 for the acceleration due to gravity and the time we found earlier, you should get an answer of about 

0.828m 
and that's your answer. 

Hope that helped.

anonymous · Answer

Thank you so much.