Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

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
```
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

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 ?

## Other Related Questions on Mechanics

View all Questions »
• Complete Physics Course - Class 12
• OFFERED PRICE: Rs. 2,756
• View Details
Get extra Rs. 2,756 off
USE CODE: CHENNA
• Complete Physics Course - Class 11
• OFFERED PRICE: Rs. 2,968
• View Details
Get extra Rs. 2,900 off
USE CODE: CHENNA