Click to Chat

1800-1023-196

+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 Rises from rest on the ground with constant acceleration 1 m per second square a stone is dropped when the balloon has Raj into a height of 39.2 metre the time taken by the stone to reach the ground nearly`
one year ago

Arun
22034 Points
```							Dear Anuj The velocity of the balloon at the height 39.2 is v = sqrt(2aH) = sqrt(2*1*39.2) = sqrt(78.4) Initial velocity of the stone at height 39.2 is u = sqrt(78.4) upwards. 39.2 = sqrt(78.4) *t - (1/2)* t²The time taken by the stone to reach the ground can be obtained by solving the above quadratic.  Hope it helps RegardsArun (askIITians forum expert)
```
one year ago
Think You Can Provide A Better Answer ?

## Other Related Questions on General Physics

View all Questions »

### 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