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

Average force will be equal to 2908.57 N

Step-by-step 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=(u^2sin2\Theta )/(g)`

`200=(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* 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=(dP)/(dt)=(20.36)/(0.007)=2908.57N`

Force will be equal to 2908.57 N