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A 0.59 kg football is thrown with a velocity of 15 m/s to the right. A stationary...
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High School Teacher
The acceleration required to stop the ball with velocity of 15m /s in 0.018s is (velocity-0)/Time to stop it = 15/0.018m/s^2.
Therefore the force exerted by ball on the catcher = ball's mass*ball's acceleration=0.59*(15/0.018) =491.6667N
Posted by neela on October 23, 2009 at 2:00 AM (Answer #1)
The initial velocity of of ball = u = 15 m/s
Final velocity of ball = v = 0
Time taken to stop the ball = t = 0.018 s
Mass of the ball = 0.59 kg
Therefore acceleration = a = (u - v)/t = -15/0.018 m/s^2
and force f = m*a = 0.59 (-15/0.018) = 491.6666 N approximately
Answer: force exerted on the receiver is 491.6666 N.
Posted by krishna-agrawala on October 23, 2009 at 2:03 AM (Answer #2)
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