9. A projectile is fired from a cliff 400 feet above the water with a muzzle velocity of 300 fee... (2025)

STEP 1

Assumptions
1. The height of the projectile above the water is given by the equation $h(x) = \frac{-32x^2}{9000} + x + 400$h(x)=9000−32x2​+x+400.
2. The variable $x$x represents the horizontal distance of the projectile from the base of the cliff.
3. The initial height from which the projectile is fired is 400 feet.
4. The acceleration due to gravity is represented as -32 feet per second squared in the equation.
5. The maximum height and the distance from the base of the cliff when the projectile hits the water are the quantities to be determined.

STEP 2

To find the maximum height of the projectile, we need to determine the vertex of the parabola represented by the equation $h(x)$h(x). The vertex form of a parabola is given by $h(x) = a(x-h)^2 + k$h(x)=a(x−h)2+k, where $(h, k)$(h,k) is the vertex of the parabola.

STEP 3

The $x$x-coordinate of the vertex can be found using the formula $x = -\frac{b}{2a}$x=−2ab​, where $a$a and $b$b are the coefficients of $x^2$x2 and $x$x in the standard form of the quadratic equation $ax^2 + bx + c$ax2+bx+c.

STEP 4

Identify the coefficients $a$a and $b$b from the given equation $h(x) = \frac{-32x^2}{9000} + x + 400$h(x)=9000−32x2​+x+400.
$$a = \frac{-32}{9000}, \quad b = 1$$a=9000−32​,b=1

STEP 5

Calculate the $x$x-coordinate of the vertex using the formula from STEP_3.
$$x = -\frac{b}{2a} = -\frac{1}{2 \cdot \frac{-32}{9000}}$$x=−2ab​=−2⋅9000−32​1​

STEP 6

Simplify the expression to find the $x$x-coordinate of the vertex.
$$x = -\frac{1}{2 \cdot \frac{-32}{9000}} = -\frac{9000}{2 \cdot -32}$$x=−2⋅9000−32​1​=−2⋅−329000​

STEP 7

Continue simplifying the expression.
$$x = \frac{9000}{64}$$x=649000​

STEP 8

Calculate the value of $x$x.
$$x = 140.625$$x=140.625

STEP 9

Now that we have the $x$x-coordinate of the vertex, we can find the maximum height by evaluating $h(x)$h(x) at $x = 140.625$x=140.625.
$$h(140.625) = \frac{-32 \cdot (140.625)^2}{9000} + 140.625 + 400$$h(140.625)=9000−32⋅(140.625)2​+140.625+400

STEP 10

Calculate the value inside the square and then multiply by $\frac{-32}{9000}$9000−32​.
$$h(140.625) = \frac{-32 \cdot 19775.390625}{9000} + 140.625 + 400$$h(140.625)=9000−32⋅19775.390625​+140.625+400

STEP 11

Simplify the expression.
$$h(140.625) = \frac{-632014.5}{9000} + 140.625 + 400$$h(140.625)=9000−632014.5​+140.625+400

STEP 12

Continue simplifying the expression.
$$h(140.625) = -70.2238333333 + 140.625 + 400$$h(140.625)=−70.2238333333+140.625+400

STEP 13

Calculate the maximum height.
$$h(140.625) = 470.4011666667$$h(140.625)=470.4011666667

STEP 14

Round the maximum height to the nearest foot.
$$h(140.625) \approx 470 \text{ feet}$$h(140.625)≈470feet
The maximum height of the projectile is approximately 470 feet.

STEP 15

To find the distance from the base of the cliff at which the projectile hits the water, we need to solve for $x$x when $h(x) = 0$h(x)=0.

STEP 16

Set the equation $h(x)$h(x) equal to zero and solve for $x$x.
$$0 = \frac{-32x^2}{9000} + x + 400$$0=9000−32x2​+x+400

STEP 17

Multiply through by 9000 to clear the fraction.
$$0 = -32x^2 + 9000x + 3600000$$0=−32x2+9000x+3600000

STEP 18

Use the quadratic formula to solve for $x$x, where $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$x=2a−b±b2−4ac​​.

STEP 19

Identify the coefficients $a$a, $b$b, and $c$c from the equation in STEP_17.
$$a = -32, \quad b = 9000, \quad c = 3600000$$a=−32,b=9000,c=3600000

STEP 20

Calculate the discriminant $\Delta = b^2 - 4ac$Δ=b2−4ac.
$$\Delta = 9000^2 - 4 \cdot -32 \cdot 3600000$$Δ=90002−4⋅−32⋅3600000

STEP 21

Simplify the discriminant.
$$\Delta = 81000000 + 460800000$$Δ=81000000+460800000

STEP 22

Calculate the value of the discriminant.
$$\Delta = 541800000$$Δ=541800000

STEP 23

Now, use the quadratic formula to find the values of $x$x.
$$x = \frac{-9000 \pm \sqrt{541800000}}{2 \cdot -32}$$x=2⋅−32−9000±541800000​​

STEP 24

Calculate the square root of the discriminant.
$$\sqrt{541800000} \approx 23277.649$$541800000​≈23277.649

STEP 25

Calculate the two possible values for $x$x.
$$x_1 = \frac{-9000 + 23277.649}{-64}, \quad x_2 = \frac{-9000 - 23277.649}{-64}$$x1​=−64−9000+23277.649​,x2​=−64−9000−23277.649​

STEP 26

Simplify the expressions to find the two possible values for $x$x.
$$x_1 = \frac{14277.649}{-64}, \quad x_2 = \frac{-32277.649}{-64}$$x1​=−6414277.649​,x2​=−64−32277.649​

STEP 27

Calculate the values of $x_1$x1​ and $x_2$x2​.
$$x_1 \approx -223.089, \quad x_2 \approx 504.339$$x1​≈−223.089,x2​≈504.339

STEP 28

Since a negative distance does not make sense in this context, we discard $x_1$x1​ and take $x_2$x2​ as the distance from the base of the cliff at which the projectile hits the water.
$$x = 504.339 \text{ feet}$$x=504.339feet

STEP 29

Round the distance to the nearest foot.
$$x \approx 504 \text{ feet}$$x≈504feet
The projectile hits the water approximately 504 feet from the base of the cliff.