91Ó°ÊÓ

A projectile is fired from a cliff 200 feet above the water at an inclination of \(45^{\circ}\) to the horizontal, with a muzzle velocity of 50 feet per second. The height \(h\) of the projectile above the water is modeled by $$ h(x)=\frac{-32 x^{2}}{50^{2}}+x+200 $$ where \(x\) is the horizontal distance of the projectile from the face of the cliff. (a) At what horizontal distance from the face of the cliff is the height of the projectile a maximum? (b) Find the maximum height of the projectile. (c) At what horizontal distance from the face of the cliff will the projectile strike the water? (d) Graph the function \(h, 0 \leq x \leq 200\). (e) Use a graphing utility to verify the solutions found in parts (b) and (c). (f) When the height of the projectile is 100 feet above the water, how far is it from the cliff?

Step by step solution

01

Find the vertex

The height function given is a quadratic equation in the form \[ h(x) = \frac{-32 x^{2}}{50^{2}} + x + 200 \].Since it is a parabola opening downwards (coefficient of \(x^2\) is negative), the maximum height occurs at the vertex, which can be found using \[ x = -\frac{b}{2a} \].Here, \(a = \frac{-32}{50^2} = -\frac{32}{2500}\) and \(b = 1\).So, \[ x = -\frac{1}{2 \times -\frac{32}{2500}} = \frac{2500}{64} = 39.06. \]Therefore, the horizontal distance at which the height is maximum is approximately 39.06 feet.

02

Maximum height of the projectile

Substitute \(x = 39.06\) into the height function to find the maximum height:\[ h(39.06) = \frac{-32(39.06)^2}{50^2} + 39.06 + 200. \]Calculate the terms:\[ h(39.06) = \frac{-32(39.06)^2}{2500} + 39.06 + 200 = -19.54 + 39.06 + 200. \]Therefore, the maximum height is approximately 219.52 feet.

03

Horizontal distance when the projectile strikes the water

Set the height function to zero and solve for \(x\):\[ 0 = \frac{-32 x^{2}}{50^{2}} + x + 200. \]Multiply through by 2500 to clear the fraction:\[ 0 = -32x^2 + 2500x + 500000. \]Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \(a = -32\), \(b = 2500\), and \(c = 500000\).Calculate the discriminant:\[ b^2 - 4ac = (2500)^2 - 4(-32)(500000) = 6250000 + 64000000 = 70250000. \]Solve for \(x\):\[ x = \frac{-2500 \pm \sqrt{70250000}}{-64}. \]Approximating the root and then solving,\[ x \approx \frac{11454.13}{64} = 178.97. \]Thus, the projectile strikes the water at approximately 178.97 feet from the cliff.

04

Graph the function

Graph the function using the given height formula:\[ h(x) = \frac{-32 x^{2}}{50^{2}} + x + 200, \]in the interval \(0 \le x \le 200\). Plot the points and draw the parabola.

05

Verify with a graphing utility

Use a graphing utility to plot the function and verify the solutions for the maximum height and the horizontal distance where the projectile strikes the water.

06

Horizontal distance when height is 100 feet

Set the height function equal to 100 and solve for \(x\):\[ 100 = \frac{-32 x^{2}}{50^{2}} + x + 200 \].Rearrange to form a quadratic equation:\[ 0 = \frac{-32 x^{2}}{2500} + x + 100. \]Multiply through by 2500:\[ 0 = -32x^2 + 2500x + 250000. \]Solve using the quadratic formula where \(a = -32\), \(b = 2500\), and \(c = 100000\).Calculate the discriminant:\[ b^2 - 4ac = (2500)^2 - 4(-32)(100000) = 6250000 + 12800000 = 19050000. \]Solve for \(x\):\[ x = \frac{-2500 \pm \sqrt{19050000}}{-64}. \]Approximating the root values and solving,\[ x_1 \approx 2.881 \text{ and } x_2 \approx 108.57. \]Therefore, the horizontal distances when the height is 100 feet are approximately 2.88 feet and 108.57 feet.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Equations

Quadratic equations are a type of polynomial equation that can be written in the standard form: \[ ax^2 + bx + c = 0 \].They are characterized by the term with the variable raised to the second power (i.e., \(x^2\)). In projectile motion problems, these equations help model parabolic trajectories. For the given exercise, the height function of the projectile is modeled by a quadratic equation:\[ h(x) = \frac{-32 x^2}{50^2} + x + 200 \].This formula describes how the height of the projectile changes with horizontal distance (\(x)\) from the cliff. The coefficients determine the shape and direction of the parabola in such models. Understanding this helps solve for maximum height, range, and points of impact.

Vertex Formula

The vertex formula is used to find the maximum or minimum point of a quadratic equation. For a parabola given by \[ y = ax^2 + bx + c \], the x-coordinate of the vertex (where the maximum or minimum value occurs) can be found using:\[ x = -\frac{b}{2a} \].In our exercise, the coefficients are \( a = \frac{-32}{50^2} \approx -0.0128 \) and \( b = 1 \). Substituting these values, we get:\[ x = -\frac{1}{2 \times -0.0128} \approx 39.06 \].This means the maximum height occurs at 39.06 feet from the cliff. Understanding the vertex formula guided us to correctly determine where the projectile reaches its peak height.

Parabolic Graphing

Graphing parabolas is crucial to visualize the path of the projectile. In our function: \[ h(x) = \frac{-32 x^2}{50^2} + x + 200 \],we recognize it is a downward-opening parabola (since the coefficient of \(x^2\) is negative). To graph this:

• Find key points like the vertex (maximum height at 39.06 feet horizontally).
• Calculate the value of the function at key points.
• Plot these points and draw a smooth curve passing through them.

We plotted points up to a horizontal distance of 200 feet (from given interval 0 ≤ x ≤ 200), showing a clear parabolic path that verifies our mathematical determinations of the maximum height and range.

Discriminant

The discriminant of a quadratic equation \[ ax^2 + bx + c = 0 \] is given by \[ \Delta = b^2 - 4ac \].This value helps determine the nature of the roots:

• If \( \Delta > 0 \), the quadratic equation has two distinct real roots.
• If \( \Delta = 0 \), it has exactly one real root (a repeated root).
• If \( \Delta < 0 \), the roots are complex and not real.

In solving our exercise's height function to find where the projectile strikes the water,we set the equation to zero:\[ 0 = \frac{-32x^2}{50^2} + x + 200 \] and solved \[ 0 = -32x^2 + 2500x + 500000 \]with discriminant:\[ \Delta = 2500^2 - 4(-32)(500000) = 70250000 \].A positive discriminant means two distinct real roots, indicating two different horizontal distances (0 and 178.97 feet) where the height matches a specified value.Understanding the discriminant assists in predicting the behavior and intersections of quadratic graphs.