Question

In the figure, a baseball is hit at a height h = 1.10 m and then caught at the same height. It travels alongside a wall, moving up past the top of the wall 1.3 s after it is hit and then down past the top of the wall 3.4 s later, at distance D = 55 m farther along the wall. (a) What horizontal distance is traveled by the ball from hit to catch? What are the (b) magnitude and (c) angle (relative to the horizontal) of the ball's velocity just after being hit? (d) How high is the wall?

Answer 1

First the total horizontal distance= Vox*T (T=total time)
Well Vox is a constant, Vox=55m/3.4s=16.17m/s and total time is 4.7 seconds (3.4 seconds above the wall and 1.3 seconds next to).

Then to find the angle:
you know that the ball was traveling 4.7 seconds total, so you know (because the ball ends at the same height) that half the time (2.35s) is where the ball is at the highest point. And you know:
Vy=Voy-9.8t and Voy is zero at the top so you solve for Voy by plugging in the time (2.35s) and Vy (O). then calculate the height with:
Y=Yo-Voyt-4.9t^2 with t=2.35s.

So take how high the ball goes (Y-1) and the total horizontal distance (half of it anyway) and use that triangle to tan(theta)=(Y-1)/(X/2) and solve for theta.

And the wall's height you use Y=Yo-Voyt-4.9t^2 and plug in t=1.1 to find out how high the wall is (you already solved Voy, Yo=1.10m).