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: 11
        a balloon is ascending vertically with an acceleration of 0.2 m/s2. Two stones are dropped from it at an interval of 2sec. Find the distance between them 1.5 sec after the second stone is released. (use g=9.8)
7 years ago

Answers : (1)

gauhar singh
7 Points
							

first u have to specify the velocity of balloon when stone is dropped,let it be v,

velocity of 1st stone is v in upward direction n acceleration downward, use

vt - 0.5 gt2 = d using appropriate signs for directions. t=2+1.5 sec

the 2nd stone will have addition separation due to movement of balloon n also additional speed

use same formula with t=1.5 sec and v2 = v + time.acceleration of balloon

*make a diagram with reference point as the place where 1st stone is drawn n measure distances accordingly

7 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

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


Course Features

  • 18 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