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. -16x^2+129x+119

Answer 1

Answer:

8.90 seconds

Step-by-step explanation:

When the rocket hits the ground, its height ( y ) will be zero. This means we can write the equation as:

• 0 = -16x²+129x+119

So we need to solve that equation:

• a = -16

• b = 129

• c = 119

• x = $\frac{-b\frac{+}{-}\sqrt{b^2-4ac} }{2a}$

x₁ = -0.84

x₂ = 8.90

We discard x₁, as it is negative.

So the answer is x₂, 8.90 seconds.