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Projectile Motion ball thrown

  1. Jun 18, 2012 #1
    A ball A is released from rest at a height of h and another ball B is provided with a horizontal force at the same height h. Both the balls fall to the ground. Which ball will reach the ground first?

    Doesn't ball A reach the ground first because the ball B is applied with a horizontal force and so will remain in air for a longer time...???

    Do they both fall to the ground at the same time??
  2. jcsd
  3. Jun 18, 2012 #2
    They both fall to the ground at the same time. A force applied in the horizontal direction doesn't affect its motion in the vertical direction.
  4. Jun 19, 2012 #3
    yes but how and why..... doesn't it defy conventional wisdom that it will be away from the ground for a longer time?
  5. Jun 19, 2012 #4
    Because the ball that goes away has initial "away" speed. Time for both are equal but the one with "away" property will perform better. It has added capability to go down and away.
  6. Jun 19, 2012 #5
    What do you mean by:-

    I don't understand....
  7. Jun 19, 2012 #6
    I guess your understanding is that the body has to go down to the ground AND then move forward. So your reasoning that it need more time for 2 motions.

    The body can do 2 jobs at a time.
    In 1 sec, it can go down and that within that interval also it goes forward.
    So if you find a value that it goes down, that interval also it does another job, moving forward.

    I call it Multitasking.
  8. Jun 19, 2012 #7
    k thanks..got what you meant!!!! :)
  9. Jun 19, 2012 #8


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    If you really wanted to you could write equations for the distance the two balls travel through the air and their velocity. The one given a horizontal push obviously travels further through the air (in a curved path) but it also moves faster through the air due to the extra force acting on it. If you then worked out Time = distance/velocity you would find the extra velocity and extra distance "cancel" so the total time of flight is the same and both balls hit the ground together (ignoring air resistance).

    However the maths is a lot easier if you realise that you can work out the vertical and horizontal components separately! This trick can even be used where the horizontal push isn't exactly horizontal. In that case you would work out what the vertical and horizontal components of the push were and then solve the equations for vertical and horizontal motion separately. Obviously in this case the balls would not reach the ground at the same time.

    The classic school homework question involves a cannon pointed up at some angle and asks you to work out how far the ball will go. Again the way to approach it it to write separate equations for the vertical and horizontal motion. Remembering that as the ball reaches peak altitude the vertical component of it's velocity is zero and that the flight time both vertically and horizontally is the same.
    Last edited: Jun 19, 2012
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