# A 0.59 kg football is thrown with a velocity of 15 m/s to the right. A stationary receiver catches the ball and brings it to rest in 0.018 s.What is the force exerted on the receiver? Answer in...

A 0.59 kg football is thrown with a velocity of 15 m/s to the right. A stationary receiver catches the ball and brings it to rest in 0.018 s.

What is the force exerted on the receiver? Answer in units of N.

Asked on by hrrjack

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on

Given

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.

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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

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