posted by shan .
A baseball bat makes contact with a ball 1.05 meters above the ground and the ball moves away at a 41.5 degree angle from the horizontal. It's a homer!! The ball lands in the bleachers 133.5 meters away from home plate, 4.1 meters above the ground. How long was the ball in the air? How fast was the ball going when it left the bat?
Let T be the time the ball is in the air.
V cos 41.5 * T = 133.5
V sin 41.5 * T - (g/2) T^2 + 1.05 = 4.1
Now solve these two simulataneous equations for V and T. The first eqaution tells you that V = 178.2/T
Substitute that for V in the second equation to eliminate one of the two variables