### Math Notes

Subjects

#### Algebra Solutions

##### Topics || Problems

If an object is shot vertically from the surface of the earth with an initial velocity of v feet per second and if the 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 feet per second square. (a) Solve t in terms of s. (b) If v = 300 feet per second use (a) to find when s =450 and s= 0. Solution:

(a) Solve t in terms of s.

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

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

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

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

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

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

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

If $$v =300 \text{fps}$$ solve $$t$$ at $$s = 450$$

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

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

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

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

If $$v =300 \text{fps}$$ solve $$t$$ at $$s = 0$$

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

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

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

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