Question

an inquisitive physics student and mountain climber climbs a 50 m high cliff that overhangs a calm pool of water. he throws two stones vertically downward, 1 s apart and observes that they cause a single splash. the first stone has an initial speed of 2 m/s. How long after release of the first stone do the two stones hit the water?

Answer 1

initial velocity=v0=2m/s
displacement=h-h0=-50m
acc due to gravity=g=9.8m/s^2
time taken by the stones to hit the water=t
Calculate 't' using
h-h0=(v0)t-(1/2)gt^2.

Answer 2

oh lol of course..what about the initial velocity of the second stone if the two stones hit the water simultaneously? i would equate the distances right?

Answer 3

or do i equate the time?

Answer 4

both.

Answer 5

i see i see thanks :) i guess i can go from here on

Answer 6

yw.