A mountain climber stands at the top of a 70.0m cliff hanging over an ocean. The climber throws thwo stones vertically 1.5s apart and observes that they cause a single splash when they hit the water. The first stone has an initial velociy of +2.0m/s
How long after the reliease of the first stone will the two stones hit the water?
What is the initial velocityt of the second stone?

I have no idea how to do this! Please help

Let initial velocity be u.

s = ut + 1/2 ft^2 (f = acceleration) (Equation A)

In case of the first stone, the above is:

70 = 2t +1/2 g t^2 (g = 9.81) Eqauation C

Find value of t from above.

t =3.57

For the second stone, time = t - 1.5 =2.07

From equation A

70 = u 2.07 + 1/2 g 2.07^2

u = 24 (Approx)

ANSWER:

Initial velocity of second stone = 24
Time = 3.57

Set up a table for both of the stones. (This table works for any free-fall problem by the way)

stone 1:
y(initial)= 70 m (since it starts at the top of the cliff)
y(final)= 0 m (since it ends at the bottom of the cliff)
v(initial)= +2 m/s (Since it is a + velocity it means it was thrown upward)
v(final)=?
a= -9.8 m/s^2 (this is always true near the Earth's surface and the object is in free-fall)
t= t1 (some unknown time, but I will label it t1)

stone 2:
y(initial)= 70 m (since it starts at the top of the cliff)
y(final)= 0 m (since it ends at the bottom of the cliff)
v(initial)=?
v(final)=?
a= -9.8 m/s^2 (this is always true near the Earth's surface and the object is in free-fall)
t= t1-1.5s (since stone 1 is thrown first, and both stones hit the water at the same time, then the second stone must be in the air 1.5s less than the first stone.)


Now for each stone you can use your kinematic equations to solve for your unknown stuff. You should be able to solve for all the stuff for stone 1 first, then move on to stone 2

These tables are a good way to organize the information given to you.