|

Mechanics

New kid on the Block

 Joined: 14 May 2009 Post: 4
14 May 2009 15:49:23 IST
0 People liked this
13
620
A balloon ascends with a constant acceleration of g/8.At the end of 30 sec a body is released from b
Engineering Entrance , Medical Entrance , NEET , JEE Main , AIIMS , JEE Main & Advanced , Physics , Mechanics , Kinematics , NewtonsLawsCirMotion , Work Power EnergyÂ collisions , Rotation , Gravitation , Simple Harmonic Motion , Fluid Mechanics , Materials Wave Sound , Super Position , units dimensions , com momentum , shm , wave sound

A balloon ascends with a constant acceleration of g/8.At the end of 30 sec a body is released from balloon. Find time taken by body to reach ground?

Cool goIITian

Joined: 7 May 2009
Posts: 51
14 May 2009 16:09:15 IST
0 people liked this

ans: 15/4

at the end of 30 sec speed of baloon is v=30g/8(using v=u+at formula)

so speed of thhe body released is also 30g/8in the upward direction

the body will fall with normal acc due to gravity i.e g

0=-30g/8+gt

t=15/4

cheers...

Cool goIITian

Joined: 4 Jul 2008
Posts: 82
14 May 2009 16:10:12 IST
0 people liked this

Cool goIITian

Joined: 4 Jul 2008
Posts: 82
14 May 2009 16:10:23 IST
0 people liked this

tell me is this correct?

Blazing goIITian

Joined: 25 Jan 2009
Posts: 311
14 May 2009 16:13:17 IST
2 people liked this

wolverine is absolutely wrong ...the velocity on reaching the ground will be more than 0

Blazing goIITian

Joined: 30 Mar 2008
Posts: 1418
14 May 2009 16:15:12 IST
0 people liked this

vel at the end of 30 sec is 30g/8

and the height attend is 1125/2 m

now apply formula s=ut+1/2at^2....

Blazing goIITian

Joined: 30 Mar 2008
Posts: 1418
14 May 2009 16:20:44 IST
0 people liked this

Blazing goIITian

Joined: 30 Mar 2008
Posts: 1418
14 May 2009 16:21:03 IST
0 people liked this

tell me is the answer is corrct

Blazing goIITian

Joined: 9 May 2009
Posts: 831
14 May 2009 16:32:53 IST
1 people liked this

when stone is released from balloon its height is h=1/2at^2 = 1125/2

and  vel 30g/8

time taken by stone to reach the ground..

t =

apply this formula and get the result

Blazing goIITian

Joined: 9 May 2009
Posts: 831
14 May 2009 16:33:37 IST
0 people liked this

if i m wrong tell me

Blazing goIITian

Joined: 25 Jan 2009
Posts: 311
14 May 2009 16:49:54 IST
1 people liked this

at the end of 30 sec the velocity of balloon is 300/8  and distaance travelled is 9000/16

now use the eqution-----

9000/16=   -300/8t  +1/2 g t^2

simplified equation-----2t^2   -15t  -225=0  solving this we get 15 sec as ans

Blazing goIITian

Joined: 25 Jan 2009
Posts: 311
14 May 2009 16:51:25 IST
0 people liked this

guys tell me if i m rit or wrong

Blazing goIITian

Joined: 30 Mar 2008
Posts: 1418
14 May 2009 17:01:00 IST
0 people liked this

okay the answer is 15...rafa is correct...

Forum Expert
Joined: 19 Oct 2006
Posts: 7802
15 May 2009 17:21:35 IST
1 people liked this

Q) A balloon ascends with a constant acceleration of g/8.At the end of 30 sec a body is released from balloon. Find time taken by body to reach ground?

Sol) t = 30s, u = 0, a = g/8

thus using S = ut + 1/2 a*t2

S = 0.5 * g/8 * 900 = 551.25 m

Moreover v = u + at = g/8 * 30 = 15g/4 m/s is the velocity of the body in upward direction when dropped from the baloon after 30s

Thus, time taken by body to fall the ground is given by

t1 + t2

where t1 = time taken by the body to come back to the point from where it was released

t2 = time taken by body to reach the gound after coming back to the point from where it was released

For Finding t1:

use v - u = at

or 15g/4 - (-15g/4) = gt1

or t1 = 30/4 = 7.5 s

For Finding t2:

Use S = ut + 0.5 a t2

or 551.25  = (15g / 4)*t + 0.5 * g * t22

Find t2 and obtain total time t1 + t2

 Some HTML allowed. Keep your comments above the belt or risk having them deleted. Signup for a avatar to have your pictures show up by your comment If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team

## For Quick Info

Name

Mobile

E-mail

City

Class

Vertical Limit

Top Contributors
All Time This Month Last Week
1. Bipin Dubey
 Altitude - 16545 m Post - 7958
2. Himanshu
 Altitude - 10925 m Post - 3836
3. Hari Shankar
 Altitude - 10050 m Post - 2205
4. edison
 Altitude - 10815 m Post - 7802
5. Sagar Saxena
 Altitude - 8625 m Post - 8064
 Altitude - 6330 m Post - 1979

Physics

Topics

Mathematics

Chemistry

Biology

Institutes

Parents Corner

Board

Fun Zone