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+242x+73

Answer 1

Answer:

t = 15.42 s

Step-by-step explanation:

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.

$y=-16x^2+242x+73$

We ned to find the time when the rocket will the ground. For this, y = 0

$-16x^2+242x+73=0$

It is a quadratic equation whose solution is given by :

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

Here, a = -16, b = 242 and c = 73

$x=\dfrac{-b+ \sqrt{b^2-4ac} }{2a}, x=\dfrac{-b-\sqrt{b^2-4ac} }{2a}\\\\x=\dfrac{-242+ \sqrt{(242)^2-4(-16)(73)} }{2(-16)}, x=\dfrac{-242- \sqrt{(242)^2-4(-16)(73)} }{2(-16)}\\\\=-0.295\ s, 15.42\ s$

Neglecting negative value.

So, it will hit the ground in 15.42 s.