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
Menu
Grade: 9
        
an object is dropped from a height of 10m. it rebounds to a height of 2.5m . if it is in contact with the floor for 0.01 sec, then what is the acceleration during contact ? 
3 years ago

Answers : (1)

Manas Shukla
102 Points
							
Taking gravity = 10m/s^2
Lets first find out the change in velocity of the ball while its in contact with the ground .
So , Let Velocity of the ball just before contact with the ground = v1
applying formula v^2=u^2 +2 a S
v1^2 = 0 + 2 x 10 x 10
v1 = root of (200) = 14.14
Let velocity of ball just after leaving the ground = v2
applying formula again v^2=u^2 +2 a S
0 = v2^2 – 2 x 10 x 2.5
v2 = root (50) = 7.07
Notice that v2 and v1 are in different directions
Now applying formula v = u + at
v2 = -v1 + a x .01
14.14 = -7.07 + .01 x a
1414 = -707 + a
acceleration = 1414 + 707 = 2121 m/s^2
3 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

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


Course Features

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


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details