IIT JEE , AIEEE Forum - Physics , Mechanics

laddu

2. please explain me very widely the concept included in :

a ballon is ascending vertically with an acc. of .2m/ssq. 2 stones r dropped from it at an interval of 2 sec.

find the distance b/w them 1.5 sec after the second stone is released.

pintug

firstly remember that mass has no effect on it time of back

sankydreams

The key to solving this prob r :
1) Loss of mass due to dropping of stone shudl be neglectd
2)The velo of balloon when the second stone was dropped will be higher than its velo when first stone was dropped
3)Gravity has no effect on accn of balloon

4)Let velo of baloon when first stone was dropped = u
Velo when sec stone was dropped = u + at = u + .2*2
These will also be the initial velo of the stones

5)Make sure to either take g as -ve or these velo as -ve.

Rate if this helps u...
Cheers!!!

catch_arnnie

a similar question was asked here, so you can follow the link http://www.goiit.com/posts/list/mechanics-motin-along-a-straight-line-path-21196.htm#121310 & see my ans. there....

iitkgp_bipin

Let the frame of reference be balloon.

When the stones are thrown their intial velocities are 0 with respect to the balloon.

Their acceleration as seen from balloon = g + 0.2 = 10

time taken by 1st stone = 2 + 1.5 = 3.5
time taken by 2nd stone = 1.5

s₁ = ut + (1/2)at² = 0 + (1/2)(10)(3.5²)
s₂ = 0 + (1/2)(10)(1.5²)

s₁ - s₂ = (1/2)(10)(3.5² - 1.5²) = 50 m

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