A man stands at the top of a 50.0 m cliff hanging over a calm pool of water. The man throws the two balls vertically 1.0s apart and observes that they cause a single splash when it hit the water. The first stone has an initial velocity of +2.0 m/s.
a) How long after relese of the first stone will the stone hit the water?
b) What is the initial velocity of the second stone when it is thrown?
c) What will the velocity of each stone be at the instant both stones hit the water?
Ok, so that is the question given, word too word so everything should be there.
For a, since the first one had a 1 s head start it had time to acclerate more and when it hits the water, it would have been farther ahead then just 1s. Therefore I do not know how to correctly figure out the exact time.
This is my attempt, but I don't know where to go from this.
vfy^2 = viy^2 + 2ay(t)
= 22 + 2(9.8)(50)
vfy = 31.368
ay = (vfy - viy) / t
t = 2.9968
I am not sure if this is even right.