A golfer hits a ball with initial velocity 53.7 m/s directed at some angle betwe

Question

A golfer hits a ball with initial velocity 53.7 m/s directed at some angle between 25.0 degrees and 35.0 degrees above the horizontal. The time of contact between the club and ball is 8.7 ms. Take the mass of the golf ball to be 46.0 g and neglect air resistance. What is the magnitude of the average force which the club applied to the ball? By how much (give magnitude only) has the ball's momentum changed 1.6 s after being hit? N kg-m/s What is the direction of the momentum change for Part (b)? Downward. Horizontally in direction ball was hit. Upward. Horizontally in direction opposite to direction ball was hit. Cannot be determined since exact direction in which ball was hit is not given.

Explanation / Answer

(a)

average force F = m*(v-u)/t

F = 0.046*(53.7-0)/(8.7*10^-3)

F = 284 N <<<==========answer

(b)

change in momnetum dP = m*g*t = 0.046*9.81*1.6 = 0.22 kgm/s

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(c)

upward

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