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: 12
        A body dropped freely from a height H travels 36 percent of total diatmce during the last second. Then H is equal to
6 months ago

Answers : (1)

Arun
12304 Points
							
Let H be the height of the building and t be the time to fall through this height H. 
Then H = ½ g t^2 = 5 t^2.  ---- (1) 
In (t-1) second the distance traveled is 1 - 0.36 H = 0.64 H. 
0.64 H = 5 (t-1)^2.    ------(2) 
(2) / (1) gives 0.64 = (t-1)^2/ t^2 
Or 0.8 = (t-1) / t 
Solving we get t= 5 second. 
Using (1) we get H = 125 m.
 
Regards
Arun (askIITians forum expert)
6 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Get Extra Rs. 276 off

COUPON CODE: SELF10


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


Get Extra Rs. 297 off

COUPON CODE: SELF10

Ask Experts

Have any Question? Ask Experts

Post Question

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