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+224x+121 y=−16x 2 +224x+121

Answer 1

Answer:

14.52 seconds

Step-by-step explanation:

equation of the distance of rocket traveled in time x is given by

$y = -16x^{2}+224 x +121$

The displacement of the rocket is zero when it hits the ground, so equate the equation equals to zero, we get

$0= -16x^{2}+224 x +121$

$16x^{2}-224 x-121=0$

$x = \frac{224\pm \sqrt{224^{2}+4\times 16\times 121}}{2\times 16}$

$x = \frac{224\pm 240.67}{32}$

x = - 0.52 second, 14.52 seconds

Time cannot have a negative value so the time taken by the rocket to hit the ground is 14.52 seconds.

Answer 2

Answer:

14.52 seconds.

Step-by-step explanation:

We have been given that the height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation $y=-16x^2+224x+121$. We are asked to find the time, when the rocket will hit the ground.

We know that the rocket will hit the ground, when height will be 0. So to find the time when rocket will hit the ground, we will substitute $y=0$ in our given equation as:

$0=-16x^2+224x+121$

Let us solve for x using quadratic formula.

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

$x=\frac{-224\pm\sqrt{224^2-4(-16)(121)}}{2(-16)}$

$x=\frac{-224\pm\sqrt{50176+7744}}{-32}$

$x=\frac{-224\pm\sqrt{57920}}{-32}$

$x=\frac{-224\pm240.66574}{-32}$

$x=\frac{-224+240.66574}{-32}, x=\frac{-224-240.66574}{-32}$

$x=\frac{16.66574}{-32}, x=\frac{-464.66574}{-32}$

$x=-0.520804375, x=14.520804375$

Upon rounding to nearest 100th of second, we will get:

$x\approx -0.52, x\approx 14.52$

Since time cannot be negative, therefore, the rocket will hit the ground after 14.52 seconds.