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 ascends with a uniform acceleration of 981 /8 cm/s2, at the end of half a minute a body is released from it, find the time that elapses before the body reaches the ground. [g = 9.81]`
7 years ago

510 Points
```										accelration of baloon is 981/8 cm/sec^2=9.8/8 m/sec^2
after half minute velocity of baloon is v=u+at
=9.8*30/8 m/sec                 (u=0)
total height attained by baloon is H=ut+at^2/2
=450*9.8/8 m
at this time the particle is dropped so its velocity will be same as of baloon in upward direction....
velocity of paticle is u=9.8*30/8 m/sec
now using s=ut+at^2/2 for particle
s=-H , u=9.8*30/8 ,g=-9.8,after putting these values we get a quadratic in t
2t^2 -15t-225=0
t=15,-7.5 sec
t=15 is required ans
```
7 years ago
Think You Can Provide A Better Answer ?

Other Related Questions on General Physics

View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details
Get extra Rs. 900 off
USE CODE: Susan900
• Kinematics & Rotational Motion
• OFFERED PRICE: Rs. 636
• View Details
Get extra Rs. 636 off
USE CODE: Susan900