# Problem finding height of a cliff

1. Sep 11, 2011

### jfeyen

1. The problem statement, all variables and given/known data

Calculate the height of a cliff if it takes 2.35 s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of 8.00 m/s

2. Relevant equations

acceleration of gravity: 9.80 m/s2 ; not sure if I need to use -9.8 instead.
the equations we've been given so far include:
v=v0+at
v2=v02+2a(x-x0)
x=x0+((v0+v)/2)t
x=x0+v0t+(1/2)at2
I'm having a hard time figuring out which formula(s) would be most helpful

3. The attempt at a solution

I think I have to find how high the rock went first so I can then subtract it from the total free-fall to the bottom of the cliff at the end to find the height of just the cliff, but I'm having difficulties knowing what to do first and where to plug things in.

If I use v2=v02+2a(x-x0) and plug in my initial velocity of 8 m/s and I'm guessing a final velocity of 0 m/s when it stops at the top before coming back down, and an acceleration of -9.8 m/s2? and an initial x of 0 I come up with a final x of 3.27 m.

Not sure if I'm on the right track at all but another place I thought about starting was using v=v0+at to find the time it took just to reach the top of the throw, right before the rock free-falls to the bottom. When I did that I put in 0 m/s for my final velocity = 8 m/s for my initial velocity + (-9.8) for the acceleration of gravity x my unknown of t and found the time for the rock to go up before starting to come back down to be .816 s. Meaning that the free-fall then took 2.35s-.816s= 1.53s.

This is where I'm now stuck. Not sure what to to with both/either of these starting points and where I'm going to be able to pinpoint just the height of the cliff. If someone could let me know if I'm on the right track at all or point me in the right direction I would be extremely grateful!!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 11, 2011

### davo789

Hi jfeyen, welcome to PF!

Your initial calculation of the height of the rock after the throw is correct. It reaches a maximum height of 3.27m above the cliff height. However, you may have come to realise that this isn't entirely useful. How else can you use this information?

Why do you think that the maximum velocity of the rock is zero at the bottom of the fall?

3. Sep 11, 2011

### jfeyen

I'm really not sure what the 3.27m means except that the height of the cliff will be 3.27m less than the total free-fall of the rock from it's maximum height. Not sure how to get from point A to point B though.
I thought the velocity at the bottom of the fall would be zero due to the fact that the rock is stopped at that point, however that may be flawed logic. Do I need to think of it more as the velocity at the moment right before it stops? I'm not sure what to do with it though.

4. Sep 12, 2011

### davo789

Let me ask you in a slightly different way: if you threw a ball in the air at 10m/s (releasing it at a height of 2m), what speed will it be travelling at when is passes 2m on its way back down?

And yes, your logic is floored as you suspect! And you have correctly corrected yourself. Essentially, you don't know what velocity it reaches the bottom at (although you can calculate it), but you shouldn't need to.

5. Sep 12, 2011

### jfeyen

Thanks for your help! I think I have it from here. What a great resource though. I'll be sure to come back with any other questions I have in physics!