# 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

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+181x+59

### 1 thought on “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”

1. The time that the rocket will hit the ground is 12.87 seconds.

### Step-by-step explanation :-

Given : The rocket is launched from a tower

$$y = – 16 {x}^{2} + 199x + 90$$

The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation.

### To find : The time that the rocket will hit the ground ?

$$\huge\bold{\textbf{\textsf{{\color{cyan}{Solution :}}}}}$$

When the rocket hit the ground i.e. height became zero y=0,

Equation is :-

$$y = – 16 {x}^{2} + 199x + 90$$

$$\bold{\textbf{\textsf{{\color{pink}{substitute}}}}} \: y = 0$$

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

$$x = – \frac{ – (199) + \sqrt{ {199}^{2} – 4( – 16)(90)} }{2( – 16)}$$

$$x = \frac{ – 199 + \sqrt{45361} }{ – 32}$$

$$x = \frac{ – 199 – \sqrt{45361} }{ – 32}$$

$$x = – 0.43 \: , \: 12.87$$