Question

Amelia launched her science fair rocket, and then backed away. The path of the rocket is given by the equation below, where y is the the height in feet amd x is the time in seconds the rocket is in the air y=-2×^2+5x+7
1)when does the rocket hit the ground?
2)what is the maximum height of the rocket?

Answer 1

Answer:

1)The rocket hit the ground at     $x = \frac{5}{4}$

2)The maximum height of the rocket = 12.468 feet

Step-by-step explanation:

Step(i):-

Given equation  

                y = -2 x² + 5 x + 7 ...(i)

Differentiating equation (i) with respective to 'x' , we get

             $\frac{dy}{dx} = -2( 2x) +5(1) = -4x +5$

Equating zero

            $\frac{dy}{dx} = 0$

⇒      -4 x +5 =0

⇒      -4 x = -5

⇒          $x = \frac{5}{4}$

The rocket hit the ground at     $x = \frac{5}{4}$

Step(ii):-

    $\frac{dy}{dx} = -4x +5$  ...(ii)

Again differentiating equation (ii) with respective to 'x' , we get

 $\frac{d^{2} y}{dx^{2} } = - 4(1) <0$

The maximum height at x = $\frac{-5}{4}$

  y = -2 x² + 5 x + 7

 $y = -2 (\frac{5}{4})^{2} + 5 (\frac{5}{4} ) + 7$

$y = \frac{-2(25)+ 25 (16)+7(64)}{64}$

$y = \frac{798}{64} = 12.468 feet$

The maximum height of the rocket = 12.468 feet