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Finding minimum V and angle given horizontal and vertical travel

  1. Sep 22, 2010 #1
    1. The problem statement, all variables and given/known data

    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

    2. Relevant 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
    3. 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 dont think it will work. Ideas?
     
  2. jcsd
  3. Sep 22, 2010 #2
    should I put this in advanced? Im not sure.
     
  4. Oct 26, 2010 #3
    Hi, im cant 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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