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14 May 2009 15:49:23 IST

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A balloon ascends with a constant acceleration of g/8.At the end of 30 sec a body is released from b

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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?

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snehil

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14 May 2009 16:09:15 IST

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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...

alex

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14 May 2009 16:10:12 IST

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answer is 1.28 sec..

alex

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14 May 2009 16:10:23 IST

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tell me is this correct?

rafa in revival mode

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14 May 2009 16:13:17 IST

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

C!-!!NMAY

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14 May 2009 16:15:12 IST

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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....

C!-!!NMAY

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14 May 2009 16:20:44 IST

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the answer is 19.21 sec

C!-!!NMAY

Blazing goIITian

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14 May 2009 16:21:03 IST

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tell me is the answer is corrct

Dasvidaniya

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14 May 2009 16:32:53 IST

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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

Dasvidaniya

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14 May 2009 16:33:37 IST

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if i m wrong tell me

rafa in revival mode

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14 May 2009 16:49:54 IST

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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

rafa in revival mode

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14 May 2009 16:51:25 IST

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guys tell me if i m rit or wrong

C!-!!NMAY

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14 May 2009 17:01:00 IST

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

edison

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Joined: 19 Oct 2006

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15 May 2009 17:21:35 IST

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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)*t₂ + 0.5 * g * t₂²

Find t₂ and obtain total time t₁ + t₂

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