Question

A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find the time that the rocket will hit the ground, to the nearest 100th of second. y=-16x^2+181x+59

Answer 1

The time that the rocket will hit the ground is 12.87 seconds.

Step-by-step explanation :-

Given : The rocket is launched from a tower

$y = - 16 {x}^{2} + 199x + 90$

The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation.

To find : The time that the rocket will hit the ground ?

$\huge\bold{\textbf{\textsf{{\color{cyan}{Solution :}}}}}$

When the rocket hit the ground i.e. height became zero y=0,

Equation is :-

$y = - 16 {x}^{2} + 199x + 90$

$\bold{\textbf{\textsf{{\color{pink}{substitute}}}}} \: y = 0$

$x = \frac{ -b + \sqrt{ {b - 4ac}^{2} } }{2a}$

$x = - \frac{ - (199) + \sqrt{ {199}^{2} - 4( - 16)(90)} }{2( - 16)}$

$x = \frac{ - 199 + \sqrt{45361} }{ - 32}$

$x = \frac{ - 199 - \sqrt{45361} }{ - 32}$

$x = - 0.43 \: , \: 12.87$