Jumping a Crevasse (Projectile Motion)

AI Thread Summary
A mountain climber jumps a crevasse horizontally with an initial speed, and the goal is to determine the height difference between the two sides based on the landing angle below the horizontal. The relevant equations include the constant horizontal velocity and the increasing vertical velocity due to gravity. The relationship between the landing angle and the velocities is expressed through the tangent function. The height difference can be derived by eliminating time from the equations of motion, leading to Δh = 1/2 * g * t^2. Clarification is provided that in the equation y = h - 1/2gt^2, y does not necessarily equal zero, as the focus is on finding the height difference Δh.
Annan
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Homework Statement


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?


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

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