A falling stone is at a certain instant 90 feet above the ground. two seconds later it is only 10 feet above the ground. if it was thrown down with an initial speed of 4 feet per second, from what height was it thrown?

The equation relevant to this is:

S (t) = So + Vot - At²/2

Therefore we can create two equations:

S (t) = 90 = So - 4t - 16.1t² - - > eqtn 1

S (t+2) = 10 = So - 4 (t+2) - 16.1 (t+2) ² - - > eqtn 2

Expanding eqtn 2:

10 = So - 4t - 8 - 16.1 (t² + 4t + 4)

10 = So - 4t - 8 - 16.1t² - 64.4t - 64.4

10 + 8 + 64.4 = So - 68.4t - 16.1t²

82.4 = So - 68.4t - 16.1t² - - > eqtn 3

Subtracting eqtn 1 by eqtn 3:

90 = So - 4t - 16.1t²

82.4 = So - 68.4t - 16.1t²

=> 7.6 = 64.4t

t = 0.118 s

Therefore calculating for initial height So:

82.4 = So - 68.4 (0.118) - 16.1 (0.118) ²

So = 90.7 ft