An arrow is shot with an initial upward velocity of 100 feet per second from a height of 5 feet above the ground. The equation h= -16t^2 + 100t + 5 models the height in feet t seconds after the arrow is shot. After the arrow passes its maximum height, it comes down and hits a target that was placed 20 feet above the ground. About how long after the arrow was shot does it hit its intended target?

Respuesta :

Since the question specified that the arrow had to reach the height only after reaching its maximum altitude, then the only correct answer is 6.096 seconds.

What is the equation referred to as?

A mathematical expression with the equals sign is referred to as an equation. Algebra is frequently used in equations. When performing calculations but not knowing the precise number, algebra is used.

Give,

[tex]20 = -16t^{2} + 100t + 5[/tex]

[tex]= > -16t^{2} + 100t - 15 = 0[/tex]

[tex]= > t = \frac{-100 ± \sqrt{100^{2} - 4(-15)(-15) } }{2(-16)} \\\\= > t = \frac{-100 ± \sqrt{9040} }{-32} \\\\Here, \sqrt{9040} = 4\sqrt{565} \\ \\= > t = \frac{-100 ± \ 4sqrt{565} }{-32} \\\\= > t = \frac{-25 ± \ sqrt{565} }{-8} \\\\= > t_{1} = 6.096 sec \\ = > t_{2} = 0.154 sec[/tex]

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