# Throwing a rock off a cliff

## Homework Statement

A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 23.0 m/s. The cliff is h = 22.0 m above a flat horizontal beach.

1. How long after being released does the stone strike the beach below the cliff?

2. With what speed and angle of impact does the stone land?

## Homework Equations

So far, I've only used this equations:
y-y0=v0y*t-.5*g*t^2

I couldn't find any other equations.

## The Attempt at a Solution

I've solved the first part already, the time the stone strikes the beach is 2.1 seconds.
But I'm still stuck on the second part, as I'm confused about which equations to use.

Try calculating the vertical speed of the rock after a time of 2.1 seconds.
Combine this vertical component of the velocity with the horizontal component given in the question. You may want to draw a quick right-angled triangle with one side equal to the horizontal component and another equal to the vertical component. The hypotenuse of this triangle will give you the speed. The angle of impact is simply the angle between the horizontal component and the vertical component.

You have to be sure that you separate your x and y components. If the rock is being thrown horizontally, v0x = 23.0 m/s and v0y = 0 m/s. See what I mean? You will have to use trig to find the final velocity (usually the inverse tangent of vy and vx).

Some other useful functions (I've written them in the x direction here, but they hold for the y direction as well):
vx = v0x + axt
vx2 = v0x2 + 2ax(x - x0)

If so, what is the angle? I couldn't find any angles at all. It says the speed and angle at impact (when it hits the ground), not at start. So how can I find them?

I still don't understand it.

The angle at which the stone was thrown was given in your initial question.

If you are throwing something horizontally, the throw is at a right angle to the ground.