Motion problem with two rocks being released at different times

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The discussion revolves around a physics problem involving two rocks released from a balloon rising at 3.0 m/s. The first rock is released from 50 meters above the ground, while the second rock is thrown downwards 2 seconds later from 56 meters. The key equation used to solve the problem is ∆d = Vi t + ½ a t², with the initial velocity of the first rock considered to be 0. There is some confusion regarding the initial velocity of the first rock, with suggestions that it should be -3 m/s due to the balloon's motion. Overall, the method applied for solving the problem is deemed acceptable, despite the initial velocity confusion.
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Homework Statement


the question is like this :you are in a balloon that is rising at a rate of 3.0m/s . you have two rocks. you release one when you are 50m above the ground. 2 seconds later you throw the other rock towards the ground . the two rocks hit the ground at the same time . what was the initial velocity of the second rock (with respect to the ground)


Homework Equations


we have to solve it only using the simple motion equations such as:
∆d = Vi t + ½ a t2



The Attempt at a Solution


what i did was to first say that :

t(rock1)=t(rock2)-2s and we also know that the second rock is released when we are 56m above the ground

then i proceeded to calculate the time it takes for the first rock to hit the ground (considering the fact that Vi for the first rock is 0)

then i subtracted 2 from this time and used it to calculate Vi for the second rock
i used ∆d = Vi t + ½ a t2 to calculate both these things

the problem is I am not sure I am doing this right and i would appreciate it if someone could confirm my method
thanks in advance
 
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that Vi for the first rock is 0
Surely -3 m/s ? The rocks will be moving with the balloon until they are dropped or thrown.

Apart from that your method looks OK.
 
Last edited:
Mentz114 said:
Surely -3 m/s ? The rocks will be moving with the balloon until they are dropped or thrown.

Apart from that your method looks OK.

oops i was thinking about a different problem
 
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