- #1

mjolnir80

- 55

- 0

## 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