Physics
posted by courtney .
A hitter hits a baseball into the air. The shortstop, who was initially 33.5 meters away, begins running directly away from home plate at 7 m/s at the moment that the ball was hit and catches the ball, while still running, at the same height from which it was hit, 3.2 seconds later. What was the veloicty of the ball as it left the bat?
This seems so easy, but I can't get it. Help!

He runs 7 m/s * 3.2 s = 22.4 m away from home plate while the ball is in the air, ending up 33.5 + 22.4 = 55.9 m away from home plate when he catches it.
You need both the vertical and horizontal componets of the velocity to get the answer. There is enough information to do that. The horizontal component Vx can be obtained from
55.9 m = Vx * 3.2 s
The vertical component Vy can be obtained from the time the ball was in the air. For 1/2 of the 3.2 s, it was decelerating; therefore
g * 1.6 s = Vy
V = sqrt [Vx^2 + Vy^2]
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