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Conservation of energy in projectiles

  1. Nov 28, 2004 #1
    an object launched vertically upward with an initial speed v. Neglect air resistance

    a)At what height h above the ground does the projectile have a speed of 0.5v? Express your answer in terms of v and the magnitude of the acceleration of gravity g.

    Using conservation of energy i found the max height (hmax) to be v^2/(2*g)

    i then came to find h at a speed at 0.5v to be (v^2/(2*g))/4. However it says i am off by a multiplicative factor. I tried to take in account the KE and PE but am confused...

    b) A ball is launched as a projectile with initial speed v at an angle (theta) above the horizontal. Using conservation of energy, find the maximum height (hmax) of the ball's flight. Express your answer in terms of v, g, and theta

    i thought it would be v^2/(2*g) and its not which makes me wonder why?

    thanks
     
  2. jcsd
  3. Nov 28, 2004 #2
    I got :
    [tex]\frac{mv^2}{2} = \frac{1}{2}*m* \frac{v^2}{2^2} + mgh[/tex]

    solve this for h...



    at the max heigth, the y-component of the velocity (=v*sin(theta)) is 0.

    [tex]\frac{1}{2}*m*v^2 = \frac{1}{2}*m*(v*cos(\theta))^2 + mgh_{max}[/tex]

    regards

    marlon
     
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