# Kinematics Assignment

## Homework Statement

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.

## Answers and Replies

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Delphi51
Homework Helper
vfy^2 = viy^2 + 2ay(t)
= 22 + 2(9.8)(50)
vfy = 31.368
Shouldn't there be a negative sign on the 9.8? Oh, and on the 50 as well. Okay, that must be correct. But I don't think the time calculation is. Perhaps a minus sign is needed there.

I would recommend good old d = Vi*t + .5*a*t² for this one, so you get the time of flight directly from the given information.

Yes, I should have used the d = Vi*t + .5*a*t², but I was too lazy to use the quadratic formula =D. And also the reason I didn't put negatives in 9.8 and 50 as well was because I considered the bottom as positive, but why would 50 be a negative as well (because 50 is the time and time can't be negative).

I will try using that equation and post what i get from it.

Thx