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When a quarterback throws a football, the throwing arm is slowed down to rest during the deceleration phase. Of the angular velocity of the throwing arm is 1800 deg/sec at the instant of ball release, then how much would the arm deceleration change if the deceleration was reduced from 2 sec to 0.5 sec

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Answer:

The deceleration is 4 times higher for 0.5sec than for 2 sec.

Step-by-step explanation:

If the deceleration happens from 1800deg/sec to zero in 2 sec we have that the magnitude is:


a_(1)=w/t_(1)=(1800deg/sec)/2sec=900 deg/sec^(2)

if now the time to decelerate is 0.5sec:


a_(2)=w/t_(2)=(1800deg/sec)/0.5sec=3600 deg/sec^(2)

The deceleration is now 4 times higher than the previous situation.

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