- #1
subhradeep mahata
- 120
- 13
- Homework Statement
- A balloon is rising upwards with a constant velocity of 5 m/s. When it is at a height of 30 m from the ground, a stone is released from it. Calculate the time taken by the stone to reach the ground.
- Relevant Equations
- s=u*t+1/2(at^2)
I solved it in ground frame like this:
-30=5t-5t2, and then solving for t which comes out to be 3 sec.(displacement is 30 m downwards)
Now, I am not getting the same answer when i am trying to solve it in the frame of the balloon.
W.r.t. balloon, the initial velocity is zero, displacement is still 30 m, and acceleration is -10 m/s2 downwards (as the balloon itself does not have any acceleration).
So, -30 = -5t2, but t is coming out to be √6 sec. But I am aware that time does not depend on frame of reference, so I should get the same answer.
Please help me out.
-30=5t-5t2, and then solving for t which comes out to be 3 sec.(displacement is 30 m downwards)
Now, I am not getting the same answer when i am trying to solve it in the frame of the balloon.
W.r.t. balloon, the initial velocity is zero, displacement is still 30 m, and acceleration is -10 m/s2 downwards (as the balloon itself does not have any acceleration).
So, -30 = -5t2, but t is coming out to be √6 sec. But I am aware that time does not depend on frame of reference, so I should get the same answer.
Please help me out.