Jumping a Crevasse (Projectile Motion)

[Annan]

Homework Statement

A mountain climber jumps a crevasse by leaping horizontally with speed v_o. 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?

Homework Equations

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

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.

 

[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.

 

[Annan]

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

 

[rl.bhat]

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

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