Home
Class 12
PHYSICS
A balloon rases up with uniform velocity...

A balloon rases up with uniform velocity 'u'. A body is dropped from ballon. The time of descent for the body is given by is

A

`sqrt(2h)/(g)`

B

`h=ut+(1)/(2)ft^(2)`

C

`h=-ut+(1)/(2)"gt"^(2)`

D

`-h=ut+(1)/(2)"gt"^(2)`

Text Solution

Verified by Experts

The correct Answer is:
C
Promotional Banner

Similar Questions

Explore conceptually related problems

A body moving with uniform velocity

A body is dropped from a height H. The time taken to cover second half of the journey is

Height of the body from the ground can be calculated by using the formula h=-ut+(1//2)(g t^2) in (a) A body projected vertically with velocity 'u' from the top of tower, reaches the ground in 't' sec. ( b) A body dropped from a balloon moving up with uniform velocity, reaches the ground in 't' sec (c) A body dropped from a helicopter moving up with uniform velocity, reaches the ground in 't' sec (d) A body projected vertically from the ground reaches the ground in 't' sec.

A packet is dropped from a balloon rising up with uniform velocity 9.8 "ms"^(-1) . If the balloon is at a height of 39.2 m from the ground at the time of dropping the stone, the stone reaches the ground after

The velocity-time graph of a body is shown in . The displacement of the body in 8 s is. .

A balloon rises with uniform velocity of 10 ms^(-1) . When the balloon is at a height of 75 m if a stone is dropped from it the time taken by it to reach the ground is

A body of mass m is moving with a uniform velocity U. A force is applied on the body due to which its velocity increases from u to v: How much work is being done by the force ?

A body falls from 80 m. Its time of descent is [g = 10 ms^(-2)]

A stone is dropped from a balloon going up with a uniform velocity of 5.0 m/s. If the balloon was 50 m high when the stone was dropped, find its height when the stone hits the ground. Take g=10 m/s^2 .

An aeroplane is moving with a velocity u. It drops a packet from a height h. The time t taken by the packet in reaching the ground will be