How many times as high to achieve twice speed

[fattydq]

A block initially at rest is allowed to slide down a frictionless ramp and attains a speed of v at the bottom. In order to achieve a speed of 2v instead at the bottom, how many times as high must the new ramp be?

What laws does this pertain to? I don't really know where to start.

 

[rl.bhat]

It pertains to the conservation of energy.
mgh = 1/2*m*v^2.

 

[fattydq]

It pertains to the conservation of energy.
mgh = 1/2*m*v^2.

But I don't understand...velocity is only on one side of that equation. So how would you work it out for this problem?

 

[Pengwuino]

Exactly! What you have is the height of an object as a function of it's speed at the end. So if you know you must be a height, h, to achieve a velocity v, what must h be in order to achieve 2v?

 

[fattydq]

Exactly! What you have is the height of an object as a function of it's speed at the end. So if you know you must be a height, h, to achieve a velocity v, what must h be in order to achieve 2v?

2h, but it says the answer is 4 times as high...that's what's confusing me...I don't know how to reach this final answer of 4.

 

[fattydq]

Nevermind, haha, I got it. Thanks a lot! Sorry for being a bit slow on the uptake : P