Question

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?

Answer 1

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