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If an object is shot vertically from the surface of the earth with an initial velocity of $$v$$ feet per second, and if air resistance and other disturbing factors are neglected, it is found that $$s = vt - 0.5gt^2$$, where $$s$$ feet is the height of the object above the surface at the end of $$t$$ seconds and $$g = 32$$, approximately. (a) Solve for $$t$$ in terms of $$s$$ (b) If $$v=300 \text{fps}$$, use (a) to find where $$s=450$$ and $$s=0$$.
Solution:
(a)

$$s = vt -0.5gt^2$$

$$2s =2vt -gt^2$$

$$gt^2 -2vt = -2s$$

$$t^2 - \frac{2v}{g}t = \frac{-2s}{g}$$

$$t^2 -\frac{2v}{g}t + (\frac{v}{g})^2 = \frac{-2s}{g} + (\frac{v}{g})^2$$

$$(t-\frac{v}{g})^2 = \frac{v^2 - 2sg}{g^2}$$

$$t - \frac{v}{g} = \frac{\sqrt{v^2-2sg}}{g}$$

$$t = \frac{v\pm\sqrt{v^2-2sg}}{g}$$

(b)

When $$s = 450$$

$$t = \frac{300 + \sqrt{300^2-2(450)(32)}}{32}$$

$$t = 17.11 \text{sec}$$

$$t = \frac{300 - \sqrt{300^2-2(450)(32)}}{32}$$

$$t = 1.64 \text{sec}$$

When $$s = 0$$

$$t = \frac{300 + \sqrt{300^2-2(0)(32)}}{32}$$

$$t = 18.75 \text{sec}$$

$$t = \frac{300 - \sqrt{300^2-2(0)(32)}}{32}$$

$$t = 0 \text{sec}$$