Question

A rocket is launched from a tower. The height of the rocket, y in feet, is rel
time after launch, x in seconds, by the given equation. Using this equation
time that the rocket will hit the ground, to the nearest 100th of second.
y = -16x² + 147x + 80

Answer 1

9514 1404 393

Answer:

  9.70 seconds

Step-by-step explanation:

The zero of the equation can be found using the quadratic formula.

  ax² +bx +c = 0

  x = (-b ±√(b² -4ac))/(2a)

We are only interested in the positive value of x, so this will be ...

  x = (-147 -√(147² -4(-16)(80)))/(2(-16)) = (147 +√2679)/32

  x ≈ 9.7028 ≈ 9.70 . . . . seconds

The rocket will hit the ground after about 9.70 seconds.