Question

An inquisitive physics student and mountain climber
climbs a 50.0-m-high cliff that overhangs a calm pool of
water. He throws two stones vertically downward, 1.00 s apart,
and observes that they cause a single splash. The first stone
has an initial speed of 2.00 m/s. (a) How long after release
of the first stone do the two stones hit the water? (b) What
initial velocity must the second stone have if the two stones
are to hit the water simultaneously? (c) What is the speed of
each stone at the instant the two stones hit the water?

Answer 1

I would start by adding the kinetic energy and the potential energy at 50.0m of the first stone, set it equal to the final kinetic energy of the same rock. Solving for v will give you final velocity for the first stone. Let me know if that makes sense.

Answer 2

Any luck?

Answer 3

I think it has something more to do with the kinematic equations... I got a but I can't get b or d for some reason.

Answer 4

Ok, well if you have t, the time it took for the first stone, then you have t-1 for the second, right? So now you need for the second stone to reach the water in t-1. You have the distance so\[s=v(t-1)+\frac{1}{2}a(t-1)^2\] My daughters are making it *extremely* difficult to concentrate at the moment, so I hope that helps.:)

Answer 5

I used conservation of energy, because using \[mgh+\frac{1}{2}mv_{i}^2=\frac{1}{2}mv_{f}^2\]will give you the speed from height pretty easily.

Answer 6

thank you! the answer right before this one helped me a lot!

Answer 7

I have learned about all that stuff yet :P We're just starting with the kinematic equations.

Answer 8

Ah, sorry for the spoiler. :) That's what I get for self-study.