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

Total Price: Rs.

There are no items in this cart.
Continue Shopping
vikas sharma Grade: 12

A BALLOON IS ASCENDING WITH AN ACC. OF 0.2m/s^2. TWO STONES ARE DROPPED FROM IT AT AN INTERVAL OF 2s.FIND THE DISTANCE BETWEEN THEM 1.5s AFTER THE SECOND STONE IS RELEASED.                                                                             

A)49m      B)48m       C)50m        D)47m

8 years ago

Answers : (1)

Badiuddin askIITians.ismu Expert
147 Points

Dear vikas

let when first stone is drop speed of the baloon is u

so after 2 sec

 h=2u + 1/2 * .2 * 4


 and after 2 sec apeed of the baloon



let h1 distance travell by first stone in (2+1.5) sec after it was dropped

 h1 =u(3.5 ) -1/2 *g *(3.5)2

let h2 distance travell by second stone in 1.5 sec

 so  h2= (u+.4)*2 -1/2 *g*(1.5)2

so different between two stone =|h1-h2-h|

                                             =50 m

Please feel free to post as many doubts on our discussion forum as you can.
 If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly.
 We are all IITians and here to help you in your IIT JEE  & AIEEE preparation.

 All the best.
Askiitians Experts

8 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete JEE Main/Advanced Course and Test Series
  • OFFERED PRICE: Rs. 15,900
  • View Details
  • Kinematics & Rotational Motion
  • OFFERED PRICE: Rs. 636
  • View Details

Ask Experts

Have any Question? Ask Experts

Post Question

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