At 10 m/s, how high is this second hill?

[deanine3]

Homework Statement

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?

Homework Equations

(a) mgh= 1/2mv^2
(b) mgh = 1/2mv^2

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.

 

[PhanthomJay]

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

 

[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.

 

[PhanthomJay]

oh, sorry, h= (1/2)(10 m/s^2)/(9.8 m/s^2)= 5.10? Does that look right?

yes, 5.1 meters

Am I supposed to divide 0.5 x 10 x 10 then/ 9.8

yes

or is it 0.5 x 10 (don't square)/9.8?

no

It seems like the 10 is squared and the 9.8 has the seconds squared.

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).