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Properties of a Free Fall

  1. Nov 19, 2014 #1
    Say a person was descending at constant rate of 1m/s while holding a ball. Then, without stopping, the person dropped the ball which assumed a free fall. Would the ball assume an initial velocity of 0m/s, or would it start at the one 1m/s that the person is traveling?

    If anyone can, please provide some sort of scientific or mathematic principle, experimental evidence, etc.

    Thank you to anyone who can try to help. :)
     
  2. jcsd
  3. Nov 19, 2014 #2

    Nugatory

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    Staff: Mentor

    The ball would start its free fall with a velocity of 1 meter/second relative to the surface of the earth, and a velocity of zero meters/second relative to the descending person. You can calculate this directly from the relationship between velocity and acceleration: ##v(t)=v_0+at## where ##v_0## is the speed when ##t=0##. The trick is that the value of ##v_0## will be different for the descending person and someone standing on the surface of the earth, which is why the word "relative" above is so important.

    That's the math. The physical intuition behind it is inertia and Newton's first law: The ball was moving at one meter per second while it was in my hand; inertia says it's still moving at that speed at the moment that my hand releases its grip, and gravity can only speed it up from there.
     
  4. Nov 19, 2014 #3
    My question is from a similar question that asks the final velocity of the ball after 2.5 seconds, and the book said that the initial velocity would be 0m/s while I and a few other students speculated that the extra velocity from the persons descent should be added to the final velocity.

    Now I see that both answers are right depending on the perspective.
    From the persons perspective: the initial velocity would be 0m/s.

    From a person on the grounds perspective: the initial velocity would be 1m/s and need to be added to the final velocity.

    Thank you Nugatory. (:
     
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