Respuesta :
Answer:
t=0.93 seconds or t=0.07 seconds
Step-by-step explanation:
If [tex]h=-16t^2+8t-1[/tex]
The time when the projectile lands on the ground is when its height, h=0.
[tex]-16t^2+8t-1=0[/tex]
Using Factoring by Perfect Squares
[tex]-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}[/tex]
Divide all through by the coefficient of [tex]t^2[/tex]
[tex]\frac{-16t^2}{-16} +\frac{8t}{-16}=\frac{1}{-16}\\t^2-\frac{1}{2}t=-\frac{1}{16}[/tex]
Next, divide the coefficient of t by 2, square it and add it to both sides.
[tex]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}\\[/tex]
Taking square roots of both sides
[tex]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}[/tex]
Therefore:
[tex]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[/tex]
