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+124x+127

Answer 1

Explanation:

so when the height is zero, that's when the rocket hits the ground. The equation will have two zeros, but one would be for a negative time, or before the rocket is launched. so we only one the one after the rocket is launched.

use the quadratic formula to solve for x ( which is time)

-124 +- sqrt [
`124^(2)` - 4*(-16)*(127) ] / 2 (-16)

-124 +- sqrt [15376 + 8128 ] / -32

-124 +- sqrt [23504 ] / -32

-124 +- 153.310 / -32

if we take the positive part it will become the negative time so skip that one try the negative part 1st

-124 - 153.310 / -32

-277.310 / -32

8.6659 seconds to hit the ground. ( time of flight) aka TOF

8.67 seconds ( rounded to nearest 100th)