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What am i doing wrong? Angle of vector?

  1. Oct 12, 2011 #1
    1. The problem statement, all variables and given/known data

    A ball is tossed so that it bounces off the ground, rises to a height of 3.10 m, and then hits the ground again 0.70 m away from the first bounce.

    2. Relevant equations
    H= .5(v^2/g) for y velocity.


    3. The attempt at a solution
    Vy=squareroot of(60.822) Vy= 7.799 m/s
    Vx= .70 m/1.59 s= .441 m/s
    inversetangent(Vy+Vx)= angle ------ is this correct?

    i did the inversetangent and it came out to be about 79.4 degrees. but that wasn't the correct answer :(
     
  2. jcsd
  3. Oct 12, 2011 #2
    your Vx , Vy are correct but the angle is wrong.

    [tex]\tan(\theta)=\frac{V_y}{V_x}=\frac{7.79}{0.44}[/tex]
     
  4. Oct 12, 2011 #3
    thank you! :)
     
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