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Dmitriy789 [7]
3 years ago
11

A golf ball with m = 46 g is struck a blow which makes an angle of 45o with the horizontal. The drive lands 200 m away on a flat

fairway. If the golf club and ball are in contact for a time of 7 ms, what is the average force of impact?
Physics
1 answer:
AfilCa [17]3 years ago
7 0

Answer:

Average force will be equal to 2908.57 N  

Explanation:

We have given mass of the ball m = 46 gram = 0.046 kg

Let velocity at which ball is projected is u m/sec

Angle at which ball is projected \Theta =45^{\circ}

Range of the ball is given R = 200 m

Range is equal to R=\frac{u^2sin2\Theta }{g}

200=\frac{u^2sin(90^{\circ})}{9.8}

u^2=1960

u = 44.27 m/sec

Change in momentum of the ball is equal to P=mu=0.46\times 44.27=20.36kgm/sec

Time of impact is given dt=7ms=0.007sec

Force is equal to rate of change of momentum

So force F=\frac{dP}{dt}=\frac{20.36}{0.007}=2908.57N

Force will be equal to 2908.57 N

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