A baseball is struck by a batter and is caught by a fielder 100 maway 5.00 s. Ig

Question

A baseball is struck by a batter and is caught by a fielder 100 maway 5.00 s. Ignore air resistance.

A. At what horizontal velocity does the ball travel on its way tothe fielder?
- I found this out to be 20.0 m/s
B. How long after being hit does the ball reach max height?
C. Determine initial velocity of the ball when first hit.
D. What is max height ball reaches in its flight?
E. What is the magnitude of the total initial velocity of the ballwhen hit?
F. At what angle to the horizontal does the ball leave the bat?

Explanation / Answer

By symmetry the ball reaches max height in 2.5 sec (parabolic arc) 0 = vy t - 1/2 g t2   y = 0after 5 sec vy = g t / 2 = 9.8 * 5 / 2 = 24.5 m/s v = (202 + 24.52) = 31.6m/s   initial speed of ball h = vy t - 1/2 g t2   where t= 2.5 sec h = 24.5 * 2.5 - 9.8 * 2.52 / 2 = 30.6m    max height reached tan = vy / vx = 24.5 /20      = 50.8 deg

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