Question

A projectile is launched from a hole in the ground one foot deep. Its height follows the equation h=-16t^2+8t-1 . Use factoring by perfect-squares to find the time when the projectile lands back on the ground.

Answer 1

Answer:

t=0.93 seconds or t=0.07 seconds

Step-by-step explanation:

If $h=-16t^2+8t-1$

The time when the projectile lands on the ground is when its height, h=0.

$-16t^2+8t-1=0$

Using Factoring by Perfect Squares

$-16t^2+8t=1\\\text{Divide all through by the coefficient of t^2}\\\frac{-16t^2}{-16} +\frac{8t}{-16}=\frac{1}{-16}\\t^2-\frac{1}{2}t=-\frac{1}{16}$

Divide all through by the coefficient of $t^2$

$\frac{-16t^2}{-16} +\frac{8t}{-16}=\frac{1}{-16}\\t^2-\frac{1}{2}t=-\frac{1}{16}$

Next, divide the coefficient of t by 2, square it and add it to both sides.

$t^2-\frac{1}{2}t+(-\frac{1}{2})^2=-\frac{1}{16}+(-\frac{1}{2})^2\\(t-\frac{1}{2})^2=-\frac{1}{16}+\frac{1}{4}\\(t-\frac{1}{2})^2=\frac{3}{16}$

Taking square roots of both sides

$t-\frac{1}{2}=\sqrt{\frac{3}{16}}\\t=\frac{1}{2} \pm \sqrt{\frac{3}{16}}\\t=\frac{1}{2} \pm \frac{\sqrt{3}}{4}$

Therefore:

$t=\frac{1}{2} + \frac{\sqrt{3}}{4} \: OR \: t=\frac{1}{2} - \frac{\sqrt{3}}{4}\\t=\frac{2+\sqrt{3}}{4} \: OR \: \frac{2-\sqrt{3}}{4}\\t=0.93seconds \: OR \: 0.07 seconds$