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 = –162+ 248x + 116a rocket is launched from a tower. The height of the rocket, y in feet, us related to the time after launch, x in seconds, by given equation. Using this equation,find the time that the rocket will hit the ground, ro the the nearest 100th of second y= -16x^2+248x+116

Answer 1

When rocket hit the ground the height of rocket from the ground is 0. So y = 0, when rocket hit the ground.

Simplify the equation -16x^2+248x+116 = 0 by using quadratic formula to obtain the value ot time, x.

$\begin{gathered} x=\frac{-248\pm\sqrt[]{(248)^2-4\cdot(-16)\cdot(116)}}{2\cdot(-16)} \\ =\frac{-248\pm\sqrt[]{68928}}{-32} \\ =\frac{-248\pm262.54}{-32} \\ =\frac{-248-262.54}{-32} \\ =-\frac{510.54}{-32} \\ =15.95 \end{gathered}$

The negative value of time is neglected as time can never be negative.

Thus time at which rocket hit the ground is 15.95 seconds.