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+258x+97

Answer 1

16.49 seconds to nearest hundredth.

Explanation:

y=-16X^2+258x+97

When the rocket hits the ground y = 0, so we have

-16X^2+258x+97 = 0

Using the quadratic formula:

x = [-258 +/- √(258^2 - 4*-16*97)]/ (2*-16)

= [-258 +/- √(72772)]/ (2*-16)

= [-258 +/- 269.763]/ (2*-16)

= 8.0625 +/- -8.430

= -0.3675, 16.4925.

We take the positive value

so the answer is 16.49 seconds.