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

given vertical velocity = vsin(theta) =3.2m/sec

given range =18m

v2sin(2theta)/g =18

2v2sin(theta)cos(theta)/9.8 =18

vcos(theta) = 27.56m/sec

tan(theta) =0.116

theta=6.63

maximum height =v2sin2(theta)/2g =0.522m

vetical velocity at peak = 0

time of flight =2vsin(theta)/g = 2 x 3.2/9.8=0.653sec

time taken to reach maximum height =vsin(theta)/g =0.326sec

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