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Basic Rectilllinear Motion questions

  1. Jun 7, 2012 #1
    this might sound like such an ammateur question but I just wanted to clarify it:

    When an object is thrown into the air we can work out the height reached from
    mgh = 0.5*mv2 (v = initial velocity)

    we cannot use v = u + 2as as the acceleration is not constant (the force given by the thrower and gravity are working in different directions)

    So far is this correct?

    At the top of the balls flight its speed is zero
    This is the big question. I am almost 100% sure this is the case because this the direction of velocity and resultant acceleration do not have to be in the same direction...therefore the only reason the ball would fall is if its final velocity (before falling) is zero

    is this correct too?
     
  2. jcsd
  3. Jun 7, 2012 #2

    Doc Al

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    Staff: Mentor

    Sure, as long as you throw it straight up.

    I think you mean v2 = u2 + 2as. Sure you can use it. The force given by the thrower is irrelevant once it leaves the hand--all you care about is the velocity after it leaves the hand.

    Note that the energy equation above also assumes constant acceleration--that gravity is the only force acting as it rises.

    See my comments above.

    Well, it's certainly true that a ball thrown straight up will have its velocity go from up, to zero at the top, then down. The acceleration is constant and always points down.
     
  4. Jun 7, 2012 #3
    thanks so much for this :) it has been good revision!!
     
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