Question

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?

Answer 1

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