You drop a ball from a window located on an upper floor of a building. It strikes the ground with speed v. You now repeat the drop, but you ask a friend down on the ground to throw another ball upward at speed v. Your friend throws the ball upward at the same moment that you drop yours from the window. At some location, the balls pass each other. Is this location.


  1. Answer:

     y = y₀ (1 – ½ g y₀ / v²)


    This is a free fall problem. Let’s start with the ball that is released from the window, with initial velocity vo = 0 and a height of the window i

              y = y₀ + v₀ t – ½ g t²

              y = y₀ – ½ g t²

    for the ball thrown from the ground with initial velocity v₀₂ = v

             y₂ = y₀₂ + v₀₂ t – ½ g t²


    in this case y₀ = 0

             y₂2 = v t – ½ g t²

    at the point where the two balls meet, they have the same height

             y = y₂

             y₀ – ½ g t² = vt – ½ g t²

             y₀i = v t

             t = y₀ / v

    since we have the time it takes to reach the point, we can substitute in either of the two equations to find the height

             y = y₀ – ½ g t²

             y = y₀ – ½ g (y₀ / v)²

             y = y₀ – ½ g y₀² / v²

            y = y₀ (1 – ½ g y₀ / v²)

    with this expression we can find the meeting point of the two balls

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