Ice cube sliding off metal roof

[Carpetfizz]

Homework Statement

Homework Equations

$$mgh = \frac{1}{2}v^2$$

The Attempt at a Solution

[/B]
I'm working on a). I tried using conservation of energy to get v.

$$mg(c+b) = \frac{1}{2}v^2$$

$$v = \sqrt{2g(c+b)}$$

After this I'm stuck. In order to get distance from knowing the velocity, we must know time which is question b). My friend said its possible to solve this problem using forces. Since there are no angles given in this problem, I'm assuming we convert sin and cos quantities into their equivalent ratios which we know? Even then, we will have the net force, but we can't get distance from acceleration without knowing the time.k

 

[TJGilb]

$mg(c+b)=\frac 1 2 v^2$​

You say you tried to use conservation of energy to get v, but I feel I should point out that your equation is wrong. If you started at rest and define zero potential at the ground, then it should be $mgh=\frac 1 2 mv^2$, in which case the m will cancel from both sides (hence why it says your answer will be independent of mass) giving you $gh=\frac 1 2 v^2$.

Edit: Nvm, your answer looks fine, it must be just a typo.

 

[TJGilb]

So, you have your velocity. You also have the direction of that velocity, you just don't realize it. Try forming a triangle out of b and d.

 

[haruspex]

So, you have your velocity. You also have the direction of that velocity, you just don't realize it. Try forming a triangle out of b and d.

@Carpetfizz has calculated the velocity at point C, not point B.

Carpetfizz, the velocity at C is not interesting. Treat the problem in two stages. First find the state of things at point B: velocity and time to get there.

 

[TJGilb]

@Carpetfizz has calculated the velocity at point C, not point B.

Carpetfizz, the velocity at C is not interesting. Treat the problem in two stages. First find the state of things at point B: velocity and time to get there.

Good catch, I didn't notice that.