Stones are thrown horizontally with the same velocity from the tops of two different buildings. One stone lands twice as far from the base of the building from which it was thrown as does the other stone. Find the ratio of the height of the taller building to the height of the shorter building.

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  1. h1/h2 = 4.9*(2t^2)/4.9*t^2
    h1/h2 = 4.9*4*t^2/4.9t^2= 4.

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  2. Correction:
    hi/h2 = 4.9*(2t)^2/4.9*t^2
    h1/h2 = 4.9*4*t^2/4.9*t^2 = 4.

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  3. Δy=Voy*t+1/2*ay*t^2

    Your looking for height so you use the Motion in Two Dimensions in the y direction. Voy is the initial velocity of zero time time (t). Since zero times anything is zero the first part of the equation is zero leaving +1/2*ay*t^2. a is a constant and can be substituted for -9.8 (approx rate of gravity. 1/2*-9.8 is -4.9 leaving -4.9*t^2. You do the same again from the building the rock landed further away but since the rock was rwice the distance this will mean twice the time so after subing in you will have -4.9*(2t)^2. To find the ratio you do -4.9*t^2/-4.9*(2t)^2 leaving you with a ratio of 4/1=4.

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