- #1

- 28

- 0

## Homework Statement

An inquisitive physics student and mountain climber climbs a 52.0-m-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 2.10 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?

first stone m/s

second stone m/s

## Homework Equations

vf^2=vi^2+2a(xf-xi)

xf=xi+vi t+1/2at^2

## The Attempt at a Solution

I attempted to find the T

xi=52m

xf=0?

vi=2.10m/s

a=-9.81m/s^2

vf= (by the first equation) = 32m/s

I tried plugging in the result in the second equation, but it's starting not to make sense.

I was told that I have to assume the stones to have been thrown vertically but with an angle, because it wouldn't make sense otherwise, but I am having difficulty getting what that angle is, and this assignment does not include the trig functions.