ok, then you know the second stone hit the water in 1.99sec after being thrown.
50=V(1.99)-4.98(1.99)^2
solve for V
Now the speed of each:
first: vf=vi+gt solve that , t=2.99
second: vf=V+gt
t=1.99
An inquisitive physics student and mountain climber climbs a 50.0m 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?
(a) X=-2t-4.9t^2
50=-2t-4.9t^2
t=2.99s
(b) X=V(t-1)-4.9(t-1)^2...
The thing that I don't understand is that the question states that both stones hit the ground at the same time (splash together), but in solving this problem we use "t-1" which would mean that the second stone hits the ground one second before the first stone. Am I interpreting the question wrong. Could you please explain fully
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