# Homework Help: Phy 111

1. Sep 10, 2008

### D4b34r5

1. The problem statement, all variables and given/known data
A stone is launched straight up by a slingshot. Its initial speed is 20.2 m/s and the stone is 1.40 m above the ground when launched. Assume g = 9.80 m/s2.

Aow high above the ground does the stone rise?
Bow much time elapses before the stone hits the ground?

2. Relevant equations
Y=Vit+ 1/2ayt2
vf2-vi2/ay=T

3. The attempt at a solution
I have tried the first formula with the time from the second and can't get the answer. I have tried everything I can think of and was even working with others at school to try and get this one. I have seen the other problem identical to this listed but I couldnt understand it. Please Help

Last edited: Sep 10, 2008
2. Sep 10, 2008

### LowlyPion

Welcome to PF.

I'm sure you meant this equation to be:
(vf2-vi2)/2ay=Y

Maybe that will be all you need?

ay = -g
Y + 1.4 = Ymax

Last edited: Sep 10, 2008
3. Sep 10, 2008

### D4b34r5

Thanks for the welcome. And yes I had that, I just input the equation wrong into the computer.

When I go through and try an answer I have gotten 2.06 for the time it takes for it to stop rising which I would think i could just input into the formula to have it be some thing like y=1.40m+20.2m/s(2.06s)+1/2(-9.8m/s2)(2.06s)2. I end up getting 22 but I dont think that is the answer, it just doesnt make sense with it rising for 2 seconds with an initial velocity of 20m/s...

4. Sep 10, 2008

### LowlyPion

Use the equation I suggested first.

That's (02 - (20.3)2)/2(-9.8) = Y
where your final velocity is 0.