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 ok mass 1kg is dropped from height 9.8m strikes with ground and rebounds at height ok 4.9m if the time ok contact between ball and ground is 0.1 sec than find impuse and average force acting on ball`
2 years ago

Arun
23742 Points
```							Dear student Velocity when it hits the ground u = sqrt(2gH) = sqrt(2 * 9.8 * 9.8) = 9.8 sqrt(2) m/s Velocity just after hitting the ground v = sqrt(2gH) = sqrt(2 * 9.8 * 4.9) = 9.8 m/s Impulse = change in momentum = mass × change in velocity = m * (v - u) = 1kg * (9.8 - (-9.8*sqrt(2)) m/s = 9.8 * (1 + sqrt(2)) kgm/s = 9.8 * 1.414 kgm/s = 13.8572 kgm/s Average Force = Impulse / Time taken = 13.8572 / 0.1 = 1.38 N RegardsArun(askIITians forum expert)
```
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