A stone is dropped from the roof of a building. 2.00 s after that, a second stone is thrown straight down with an initial speed of 30.0 m/s. It is observed that the two stones land at the same time. a) How long did it take the first stone to reach the ground

So the time it takes the first stone to fall from the top of the cliff  is 3.88 seconds.

Explanation:

We can use free fall equations to describe the motion of both stones given by

Once we have set both systems we can then solve for the time of the first stone.

Setting up free fall equations.

For the first stone, we do not know about the its initial height, nor its time, but we know it is being falling from rest so its initial velocity is 0, thus the equation that describes it will be:

Now at the bottom or ground level we will have y = 0, thus we will have

For the second stone, we know its take is 3.9 seconds past the initial time  of the first stone so we will have

which means

It has also an initial speed, but aiming downwards so . Using the given information we can write

At the bottom or ground level we have y = 0

Thus we get a system of 2 equations and 2 unknowns that are the initial height and the initial time t.

Solving the system of equations

We can solve for the height of the cliff from the first equation and plug it on the second equation.

Replacing on the second equation we get

At this point we can plug the value of the gravity and expand it in order to simplify it.

Combining like terms we get

t = 3.88s

So the time it takes the first stone to fall from the top of the cliff  is 3.88 seconds.

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s