Respuesta :
Answer:
3.20 seconds.
Step-by-step explanation:
We have been given that the equation for the acrobat's pathway can be modeled by the equation [tex]h=-16t^2+50t+4[/tex]. We are asked to find the time it take the acrobat to reach the ground.
The acrobat will reach the ground, when height (h) is 0. So we will set height equal to 0 as:
[tex]0=-16t^2+50t+4[/tex]
We will use quadratic formula to solve for t as:
[tex]t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
[tex]t=\frac{-50\pm\sqrt{50^2-4(-16)(4)}}{2(-16)}[/tex]
[tex]t=\frac{-50\pm\sqrt{2500+256}}{-32}[/tex]
[tex]t=\frac{-50\pm\sqrt{2756}}{-32}[/tex]
[tex]t=\frac{-50\pm 52.4976189936267}{-32}[/tex]
[tex]t=\frac{-50+52.4976189936267}{-32}\text{ or }t=\frac{-50-52.4976189936267}{-32}[/tex]
[tex]t=\frac{2.4976189936267}{-32}\text{ or }t=\frac{-102.4976189936267}{-32}[/tex]
[tex]t=-0.0780505935\text{ or }t=3.203050593[/tex]
[tex]t\approx -0.08\text{ or }t\approx 3.20[/tex]
Since time cannot be negative, therefore, it will take approximately 3.20 seconds to reach the ground.
