Finding minimum V and angle given horizontal and vertical travel

AI Thread Summary
To determine the minimum initial velocity (Vo) and launch angle (alpha) for a mountain climber to jump from point A to B over a crevasse, the horizontal distance is 1.8 meters, and the vertical drop is 1.4 meters. The equations of motion for horizontal and vertical travel are set up, but the challenge arises from having two unknowns without a clear third equation to solve for both Vo and alpha. Participants in the discussion suggest using the range equation, but note that it may not apply due to the elevation difference. Simplifying the equations leads to a relationship involving tangent functions, indicating a need for further assistance in solving the problem. The conversation highlights the complexities of projectile motion in non-level terrain.
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Homework Statement



a mountain climber plans to jump from A to B over a crevasse. Determine the smallest value of the climbers initial velocity Vo and the corresponding value of the angle alpha so he lands at B

Homework Equations


A to B is 1.8m horizontal distance and B is lower than A by 1.4m
Xx=Vo*T
Xy=Vo*T-.5*9.81*t^2

The Attempt at a Solution



Distance X = 1.8 = cos(alpha)*Vo*t => T=1.8/(cos(alpha)*Vo)
Distance Y = -1.4= Vo* (1.8/(cos(alpha)*Vo)-.5*9.81*(1.8/(cos(alpha)*Vo))^2

My question is I don't know what other equation to use since I have two unknown equations I thought I could use the Range equation R= (Vo^2/g)*Sin(2alpha) but this is not on level ground so I don't think it will work. Ideas?
 
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should I put this in advanced? I am not sure.
 
Hi, I am can't solve the same problem!

I did the same thing, wrote down the equation for position using unit vectors, then replaced T in Y position equation.

and simplified everything to

Y(t)= -1.4= Vo*sin(alpha)* (1.8/(cos(alpha)*Vo)-.5*32.1*(1.8/(cos(alpha)*Vo))^2

Y(t)= -1.4= 1.8Tan(alpha) -51.8*(1+tan^2 (alpha))/V^2

Please help me :)
sorry for my english
 
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