×

#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# A ball dropped on to the floor from a height of 10m rebounds to a height of 2.5m. If the ball is in contact with the floor for 0.02s , its average acceleration during contact is

2 years ago

Khimraj
3007 Points

initial velocity = $\sqrt{2gh} = \sqrt{2*10*10} = 10\sqrt{2}$
final velocity = $\sqrt{2gh} = \sqrt{2*10*2.5} = 5\sqrt{2}$
change in velocity = $10\sqrt{2} -(- 5\sqrt{2}) = 15\sqrt{2}$
average acceleration = $15\sqrt{2}/0.02 = 750\sqrt{2}$ m/s2
Hope it clears........................................
2 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Mechanics

View all Questions »

### Course Features

• 101 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution

### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions