# Homework Help: At 10 m/s, how high is this second hill?

1. Nov 11, 2008

### deanine3

1. The problem statement, all variables and given/known data
Suppose a car starts coasting from the top of a hill that is 60m high. (a)How fast will it be going at the bottom of the hill if there is no friction? (b) For the same car starting at the top of another hill and reaching the bottom, without friction, at 10 m/s, how high is this second hill?

2. Relevant equations
(a) mgh= 1/2mv^2
(b) mgh = 1/2mv^2

3. The attempt at a solution
(a) I solved as v= 34.29 m/s^2
(b) mgh-1/2mv^2 m cancel each other out, then /g making the equation
h= (1/2)(g)(v^2)
I end up with h= (1/2)(9.8 m/s^2) (10 m/s^2)= 490m. I'm not sure if I am using the information given correctly.

2. Nov 11, 2008

### PhanthomJay

You're doing fine, but you multiplied by g instead of dividing by it.

3. Nov 11, 2008

### deanine3

oh, sorry, h= (1/2)(10 m/s^2)/(9.8 m/s^2)= 5.10? Does that look right? Am I supposed to divide 0.5 x 10 x 10 then/ 9.8 or is it 0.5 x 10 (don't square)/9.8? It seems like the 10 is squared and the 9.8 has the seconds squared.

4. Nov 12, 2008

### PhanthomJay

yes, 5.1 meters
yes
no
Sometimes it's best to leave off the units when doing the math, then add them back in at the end. You have
$$h = 1/2(v^2)/g$$ which is $$h = 1/2(10)(10)/9.8 = 5.1$$
the units are $$[(m/s)(m/s)]/[m/s^2] = [m^2/s^2][(s^2)/m] = m$$(the height must be in length units, i.e. meters).