A football punter accelerates a football rest to a speed of

A football punter accelerates a football rest to a speed of 9 m/s during the time in which his too is in contact with the ball (about 0.16 s). If the football has a mass of 0.50 kg, what average force does the punter exert on the ball? After falling from a height of 29 m, 0 40-kg ball rebounds upward, reaching a height of 19 m. If the correct between ball and ground lasted 2.2 ms, what average force exerted on the ball?

Solution

Average force

               F   = m ( v _f - v_i ) / t

                    = ( 0.5 kg ) ( 9 m/s - 0 ) / ( 0. 16 s )

                    = 28.125 N

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before reaching the ground the velocity is v_i = sqrt (2 g h ) = sqrt ( 2* 9.8 m/s² * 29 m ) = 23.84 m/s

                     final velocity vf = sqrt ( 2* 9.8 * 19 ) = 19.29 m/s

average force

F = (0.45 kg )( 23.84m/s - 19.29 m/s ) / (2.2 x10^-3 s ))

        = 929.11 N