Grade 12th passGeneral Physics

A balloon starts rising from ground from rest with an upward acceleration 2m/s^2 . Just after 1 second a stone is dropped from it. The time taken by the stone to strike the ground is nearly ?

Manali Sharma

8 Years agoGrade 12th pass

6 Answers

Vivek Kumar Singh

8 Years ago

when balloon moves from rest with a constant acceleration 2 m/sec^2 for 1 sec. it travells 1 m upward and then stone dropped. At this time velocity of stone is 2 m/s upward. now Stone is moving under gravity. stone takes 1.93 sec. to strike the ground.

Santosh Datla

8 Years ago

Dear Student,

Thanks for posting your doubts. I have attached the complete solution.

Regards,
Santosh Datla
Physics Expert, askIITians.com

Santosh Datla

8 Years ago

Dear Student, Thanks for posting your doubt. I have attached the complete solution. Hope that this answers your query.

Regards,
Santosh Datla
Physics Expert,
askIITians.com

swarup kumar majee

8 Years ago

for balloon,

initial velocity of balloon=0 so,u=o

a=2m/s²

t=1s

v=u+at

v=2m/s (as u=0m/s ,a=2m/s²,t=1s)

s=ut+1/2at²

s=1m (as u=0m/s ,a=2m/s²,t=1s)

for stone,

u=-2m/s ,s=1m,a=9.8m/s²

at²+2ut-2s=0

9.8t²-4t-2=0

t=.7s

ankit singh

5 Years ago

The balloon starts to rise vertically upward with an acceleration of 2m/s^{2}
After one second (First Case)
Therefore in this case; u = 0 ; t =1; a=2
The distance traveled in this case by the formula s=ut+ \frac{1}{2}at^{2}
∴ s = 1m
Case Two:-
From there a stone was dropped...
Therefore the stone was falling under the effect of gravity
Therefore in this case; u=0;s=1;a=9.8
The time can be calculated in this case by the formula s=ut+ \frac{1}{2}at^{2}
∴ t ≈ 0.45s

ankit singh

5 Years ago

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