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Komok [63]
3 years ago
6

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 w

as thrown down with an initial speed of 4 feet per second, from what height was it thrown?
Physics
1 answer:
choli [55]3 years ago
7 0

The equation relevant to this is:

S(t) = So + Vot - At²/2 <span>

</span>

<span>Therefore we can create two equations:
<span>S(t) = 90 = So - 4t - 16.1t²                                      --> eqtn  1</span>
<span>S(t+2) = 10 = So - 4(t+2) - 16.1(t+2)²                     --> eqtn 2</span>

</span>

<span>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 
<span>82.4 = So - 68.4t - 16.1t²                                        --> eqtn 3</span></span>

 <span>
Subtracting eqtn 1 by eqtn 3:</span>

 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:<span>

<span>82.4 = So - 68.4(0.118) - 16.1(0.118)²

<span>So = 90.7 ft</span></span></span>

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Explanation:

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