A ball is dropped on to the floor from a height of 10m. It rebounds to a height of 2.5m.If the ball is in contact with the floor for 0.01s, what is the average acceleration during contact?
Asked by | 25th Sep, 2012, 08:31: PM
velocity of the ball just before contact:
u = -(20g)1/2 using energy conservarion, i.e equating PE at top with KE at bottom
velocity of ball just before leaving ground:
v = (5g)1/2
thus force imparted to ground :
F = M ( (5g)1/2 + (20g)1/2 )/0.01 the direction of force is downwards.
Answered by | 27th Sep, 2012, 09:23: AM
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Verify mobile number
Enter the OTP sent to your number