# Homework Help: Motion in One dimension: Stone

1. Oct 20, 2011

### jojo711

1. The problem statement, all variables and given/known data
A stone is thrown downward with a speed of 18 m/s from a height of 11 m. (acceleration due to gravity: 9.81 m/s2)
a) What is the speed (in m/s) of the stone just before it hits the ground?
b) How long does it take (in seconds) for the stone to hit the ground?

3. The attempt at a solution
When I attempted the first part, I used the equation V^2=Vo^2+2a(X-Xo), and plugged in the numbers, V^2=(18^2)+2(9.81)(-11) and the answer came out to be 10.401 m/s and it was wrong.
When I tried the second part, I used the equation V=Vo+at and plugged in the numbers 0=18+(9.81)t and the answer was 8.19 s.

2. Oct 20, 2011

### Staff: Mentor

Since you're taking down as negative, the acceleration must be negative.
Why are you setting V = 0? (Again, careful with signs.)

3. Oct 20, 2011

### jojo711

I got V=0 because the final velocity would be when the stone hits the mud and wouldn't that be 0?

4. Oct 20, 2011

### Staff: Mentor

They want the speed just before it hits the ground. (Once it hits the ground, the acceleration no longer equals that of gravity. So that formula would no longer apply.)

5. Oct 20, 2011

### jojo711

Oh, but how do I know what formula to use then?

6. Oct 20, 2011

### HallsofIvy

The stone's initial height was 11m. 11m above what?

7. Oct 20, 2011

### Staff: Mentor

You're using the correct formula. Once you solve the first part, then you'll have the final velocity.

Again, be careful with signs. Just to be consistent, anything that points down becomes negative. (You can also choose to call down positive. Just pick a convention and stick to it.)

8. Oct 20, 2011

### Physics697

For the second part use Vf = Vi + at
just as Doc Al said; be careful with the signs ( Vf and Vi )

You can also use this equation yf = yi + Vi(t) + 0.5at^2