A baseball is thrown from second base to first base in a double play. If the bal

Question

A baseball is thrown from second base to first base in a double play. If the ball is released with a vertical velocity of 3.2 m/s and is caught 18 m away at the same height as it had at the instant of release: a) What is its vertical velocity at the end of its flight? b) What is its vertical velocity at the peak of its flight? c) What horizontal distance has it traveled by the time it reaches the peak of its flight? d) How long does it take to get from second to first (with 0 air resistance)? e) How long does it take to get to the apex?

Explanation / Answer

a) vertical velocity at end of flight is -3.2 m/s

b) vertical velocity at peak of filght will be 0 m/s = zero

c) horizontal distance traveled during it reaches its peak is total horizontal distance traveled /2 = 18/2 =9 m

d) using equation v = u + at

t = (v-u)/a = (3.2 - (-3.2))9.8 = 6.4/9.8 = 0.653 sec

e) total time /2 = t/2 = 0.653/2 = 0.3265 sec

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