# Jumping a Crevasse (Projectile Motion)

1. Jan 24, 2010

### Annan

1. The problem statement, all variables and given/known data
A mountain climber jumps a crevasse by leaping horizontally with speed vo. If the climber's direction of motion on landing is $$\vartheta$$ below the horizontal, what is the height difference between the two sides of the crevasse?

2. Relevant equations
I'm not sure, but I think it's y=h-1/2gt2

3. The attempt at a solution
My first thought was to just manipulate the equation to give h=y+1/2gt2, but I'm sure that's not right. What's really confusing me is $$\vartheta$$, because I'm not sure how knowing that is pertinent.

2. Jan 24, 2010

### rl.bhat

Hi Annan. welcome to PF.
In the projectile motion horizontal velocity vo remains constant. But the vertical velocity increases. And it is given by v = gt......(1)
If θ is the angle of landing then tanθ = v/vo.....(2)
Difference in height Δh = 1/2*g^t^2 ........(3)
Using eq. 1 and 2, eliminate t and v and find Δh in terms of vo, g and tanθ.

3. Jan 25, 2010

### Annan

This is probably a dumb question, but in y=h-1/2gt^2, is y the final height, and therefore equal to 0?

4. Jan 25, 2010

### rl.bhat

Not necessarily. In the problem you have to find (h - y) which is Δh = 1/2*g*t^2