Hi guys,

Question

need help with this problem for physics and im having trouble visualizing it.

Kathy and mountain climber climbs a 55.0 m 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 1.91 m/s downward.
(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 they are to hit simultaneously? (downward)

(c) What is the speed of each stone at the instant the two hit the water?
first stone
m/s (downward)
second stone
m/s (downward)

Thanks for any help

Answer 1

a) solve time freefall
h=1/2 g t^2
b) h=vi*(t-1)+1/2 g (t-1)^2 you know g, t, h solve for vi
c) vf=vi+gt the first is vi=0. remember the time for the second is one second shorster.