# Finding speed of stone when it hits the ground.

## Homework Statement

A small rock is thrown upward with a speed of 30ft/s from the edge of a building 200ft above the ground. What is the speed of the rock when it hits the ground? Use acceleration due to gravity as –32ft/s squared.

## Homework Equations

integral of acceleration= v(t)
integral of v(t)= position

## The Attempt at a Solution

I know that v(t)=30, the initial position is 200 and acceleration=-32 and I know the problem is asking for the velocity of the rock, I visualize this problem under the interval [a,b] so it's asking for the velocity at b when the rock hits the ground. I just have no idea how all of the information fits together to set up an integral equation. Any help with this initial set up would be great

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andrewkirk
Homework Helper
Gold Member
Problems like this are easier to solve using energy conservation. You just equate the final kinetic energy (KE) to the initial KE plus initial potential energy (PE). You use PE = mgh where h is height and m is mass, and KE $\frac{1}{2}mv^2$.

Are you allowed to use energy?

HallsofIvy
Homework Helper

## Homework Statement

A small rock is thrown upward with a speed of 30ft/s from the edge of a building 200ft above the ground. What is the speed of the rock when it hits the ground? Use acceleration due to gravity as –32ft/s squared.

## Homework Equations

integral of acceleration= v(t)
integral of v(t)= position

## The Attempt at a Solution

I know that v(t)=30, the initial position is 200 and acceleration=-32 and I know the problem is asking for the velocity of the rock, I visualize this problem under the interval [a,b] so it's asking for the velocity at b when the rock hits the ground. I just have no idea how all of the information fits together to set up an integral equation. Any help with this initial set up would be great
You are told that the acceleration is a constant -32 ft/sec^2 so dv/dt= -32 or dv= -32dt. That should be a very easy integration. Further dx/dt= v (where x is height above the ground). Once you know v(t), that will be an easy integration.