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Find max height of inclined plane

  1. Dec 19, 2011 #1
    1. The problem statement, all variables and given/known data
    A block given an intial velocity of 2m/s up a frictionless inclined at 60 degrees to the horizontal. What is the highest point reached by the block?


    2. Relevant equations

    gsin(θ)=a
    a=v/t
    d=vit + 1/2at^2

    3. The attempt at a solution
    Found vx = vsin60= 2(sin60)=1.73

    I solved for the acceleration
    10(sin60)= 8.66 m/s^2

    Then I found time using v/t=a

    Then plugged in the time t into
    d=vit+1/2at^2

    Then used sin60= Opposite/hypotonuse to solve for the height.

    But I did not get the correct answer of .24m. Can someone tell me where I am going wrong and point me in the right direction please
     
  2. jcsd
  3. Dec 19, 2011 #2

    Simon Bridge

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    Hint: Conservation of energy.

    (note: you overcomplicated your analysis by doing too many steps - to use kinematics, you want to draw the v-t graph for the motion - it's a triangle - use this to get equations for displacement and acceleration which will be two equations with two unknowns. eg. make the final equation before you plug the numbers in. The kinematic equation you were looking for was v^2=u^2+2ad, saves wear and tear on your calculator to notice that sin(60)=(√3)/2 - watch for these special triangles.)

    [edit]
    I got it - the problem is not looking for the height the "correct" answer appears to be for the displacement along the slope.
     
    Last edited: Dec 19, 2011
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