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
zahid sharief Grade: 11
        

 

A BALLOON IS MOVING VERTICALLY UPWARD WITH CONSTANT ACCN (g/2) in upward direction. The particle A' was dropped from the balloon and 2sec later another particle B' was dropped from the same balloon. Assuming that the motion of the balloon remains unaffected find the separation distance between A' and B', 6sec after dropping particle B'. none of the particles reaches the ground during the time interval.


7 years ago

Answers : (1)

vikas askiitian expert
510 Points
										

I thnk one more parameter is needed that is the unitial velocity of baloon when particle A was dropped.....


let the velocity of baloon when particle A was dropped is Ua


2 seconds later velocity of baloon will be v=u+at


                                  Ub=Ua+g2/2= Ua+g


initial velocity of first particle A will be equal to Ua as the particle is placed on baloon and


initial velocity of second particle B will be equal to Ub


now, displacement of particle A after 8 seconds will be Sa=8Ua-64g/2


                                                                        Sa=8Ua-32g.......1


 displacement of particle B after 6 seconds is Sb=6Ub-36g/2


                                                                Sb=6(Ua+g) -18g


                                                                Sb=6Ua-12g..............2


 Sb-Sa=20g-2Ua..........


 if Ua (initial velocity of baloon) is known then Sb-Sa can be calculated......


 

7 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete Physics Course - Class 12
  • OFFERED PRICE: Rs. 2,756
  • View Details
  • Complete Physics Course - Class 11
  • OFFERED PRICE: Rs. 2,968
  • View Details

Ask Experts

Have any Question? Ask Experts

Post Question

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