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Maximum Height Obtained?

  1. Nov 24, 2009 #1
    Hello! My question is...What is the maximum height obtained by a 125 g apple that is slung from a slingshot at an angle of 78 degrees from the horizontal with an initial velocity of 18 m/s?

    Thanks!
     
  2. jcsd
  3. Nov 24, 2009 #2
    Any attempts/ideas on your part? This has to do with conservation of energy.
     
  4. Nov 24, 2009 #3
    I thought I could use .5Vo^2/g but I don't know how to take mass into account =\
     
  5. Nov 24, 2009 #4
    In an ideal situation without air resistance ect. all mass undergoes same downward acceleration of g near the surface of the earth. Hence, mass is not used in this question.

    Start by finding the vertical component of the apple velocity
     
  6. Nov 24, 2009 #5
    E1 = E2

    The apple is still moving at the top of the arc, so there's still some kinetic energy left over.

    KE1 = KE2 + PE2

    (1/2) * m * (v1)^2 = (1/2)*m*(v2)^2 + m*g*h2

    At the peak of the arc, there is no y-component of velocity. All that's left at the top is the x-component. Since there is no acceleration in the x direction, the x-velocity of the apple at the top will be the same as it was at the beginning

    vx1 = vx2 = v2 = v1*cos(78)


    Plug v2 into energy equation.

    (1/2) * m * (v1)^2 = (1/2)*m*(v1*cos(78))^2 + m*g*h2

    Optional step: factor "m" out of every term. The "m"s drop and you're just left with...
    (1/2)*(v1)^2 = (1/2)*(v1*cos(78))^2 + g*h2



    Solve for h2

    h2 = [ (1/2)*m*(v1)^2 - (1/2)*m*(v1*cos(78))^2 ] / (m*g)

    = ((1/2)*0.125kg*(18m/s)^2 - (1/2)*0.125kg*(18m/s*cos(78degrees))^2)/(0.125kg*9.80665m/s^2)


    h2 = 15.805314 meters

    Hope this helps!

    -Alex

    http://www.alexpleasehelp.com
     
  7. Nov 25, 2009 #6
    Thank y9u all!!
     
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