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.


EQUATION:

y=-16x^2+281x+137

Answer 1

Answer:

It will take 18.04s for the rocket to hit the ground.

Step-by-step explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$.

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = (x - x_{1})*(x - x_{2})$, given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this problem:

Height given by the following equation:

$y = -16x^{2} + 281x + 137$

It hits the ground when y = 0. So

$-16x^{2} + 281x + 137 = 0$

Multiplying by -1

$16x^{2} - 281x - 137 = 0$

So $a = 16, b = -281, c = -137$

Then

$\bigtriangleup = (-281)^{2} - 4*16*(-137) = 87729$

$t_{1} = \frac{-(-281) + \sqrt{87729}}{2*16} = 18.04$

$t_{2} = \frac{-(-281) - \sqrt{87729}}{2*16} = -0.4747$

It cannot take negative time, so we discard $t_{2}$

It will take 18.04s for the rocket to hit the ground.