Question

A projectile is fired at an inclination of \(45^{\circ}\) to the horizontal, with a muzzle velocity of 100 feet per second. The height \(h\) of the projectile is modeled by $$h(x)=\frac{-32 x^{2}}{100^{2}}+x$$ where \(x\) is the horizontal distance of the projectile from the firing point. (a) At what horizontal distance from the firing point is the height of the projectile a maximum? (b) Find the maximum height of the projectile. (c) At what horizontal distance from the firing point will the projectile strike the ground? (d) Graph the function \(h, 0 \leq x \leq 350\). (e) Use a graphing utility to verify the results obtained in parts (b) and (c). (f) When the height of the projectile is 50 feet above the ground, how far has it traveled horizontally?

Short answer

(a) 156.25 feet (b) 78.125 feet (c) 312.5 feet (d) Graph the function from 0 to 350. (e) Verify using graphing utility (f) 93.75 or 218.75 feet.

Step-by-step solution

Find the horizontal distance for maximum height

The height function is given by \[h(x) = \frac{-32 x^2}{100^2} + x\]. Rewrite the function as \[h(x) = \frac{-32 x^2}{10000} + x = -\frac{32}{10000} x^2 + x\].This is a quadratic function of the form \(ax^2 + bx + c\) where \(a = -\frac{32}{10000}\) and \(b = 1\). The maximum height occurs at \(x = -\frac{b}{2a}\). Substituting \(a\) and \(b\) in, we get: \[x = -\frac{1}{2(-\frac{32}{10000})} = \frac{10000}{64} = 156.25\] feet.

Find the maximum height

Substitute \(x = 156.25\) into the height function\[h(156.25) = \frac{-32 (156.25)^2}{100^2} + 156.25\]Calculating further,\[h(156.25) = \frac{-32 (156.25)^2}{10000} + 156.25 = -78.125 + 156.25 = 78.125\] feet.

Determine where the projectile strikes the ground

The projectile strikes the ground when the height is zero. Set the height function equal to zero and solve for \(x\):\[\frac{-32 x^2}{100^2} + x = 0\]Factor out \(x\):\[x \left(-\frac{32 x}{100^2} + 1\right) = 0\]Solve for \(x\):\[x = 0\] or \[1 - \frac{32 x}{10000} = 0 \Rightarrow x = 312.5\] feet.The projectile strikes the ground at \(x = 312.5\) feet.

Graph the function

Graph the function \[h(x) = \frac{-32 x^2}{100^2} + x\] for the range \(0 \leq x \leq 350\). The graph is a downward-opening parabola with its vertex at (156.25, 78.125).

Verify results using a graphing utility

Use a graphing utility to input the function \[h(x) = \frac{-32 x^2}{100^2} + x\] and verify the maximum height at \(x = 156.25\) feet is 78.125 feet and that the projectile strikes the ground at \(x = 312.5\) feet.

Find horizontal distance for height 50 feet

Set the height function to 50 feet and solve for \(x\):\[50 = \frac{-32 x^2}{100^2} + x\]Rearranging and solving the quadratic equation gives:\[0 = -\frac{32 x^2}{100^2} + x - 50\]Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = -\frac{32}{10000}\), \(b = 1\), and \(c = -50\), we get:\[x = \frac{-1 \pm \sqrt{1^2 - 4(-\frac{32}{10000})(-50)}}{2(-\frac{32}{10000})} = \frac{-1 \pm \sqrt{.36}}{-.0064} = \frac{-1 \pm .6}{-.0064} = 93.75\] or \(218.75\) feet.

Key concepts

Quadratic Functions

A quadratic function is a type of polynomial function where the highest degree of the variable is 2. The standard form of a quadratic function is written as \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. In the context of projectile motion, the height of the projectile as a function of its horizontal distance often forms a parabolic curve described by a quadratic function. Given the height function \( h(x) = \frac{-32 x^2}{100^2} + x \), we can see its quadratic nature as it adheres to the form \( ax^2 + bx + c \) with \( a = \frac{-32}{10000} \) and \( b = 1 \). This quadratic function models the trajectory of the projectile.

Understanding quadratic functions helps us predict and analyze the motion of objects under gravity. They provide information about the maximum height, time of flight, and the distance traveled.

Vertex Form

The vertex form of a quadratic function provides a clear way to identify the maximum or minimum point, known as the vertex, of the parabola. The vertex form is given as \( a(x-h)^2 + k \), where \( (h, k) \) represents the coordinates of the vertex.

To find the vertex in a standard quadratic function like \( h(x)=\frac{-32 x^{2}}{10000}+x \), we use the formula:
\( x = -\frac{b}{2a} \). For our function, substituting \( a \) and \( b \) gives:\( x = \frac{10000}{64} = 156.25 \).
This x-value is the horizontal distance at which the projectile reaches its maximum height. To find the height at this distance, substitute \( x \) back into the function:
\( h(156.25) = -\frac{32}{10000} (156.25)^2 + 156.25 = 78.125 \). The vertex is thus \( (156.25, 78.125) \), indicating the maximum height of the projectile is 78.125 feet at a distance of 156.25 feet from the firing point. This approach simplifies finding key points on the projectile's path.

Solving Quadratic Equations

One common method for solving quadratic equations, especially when dealing with projectiles, is using the quadratic formula:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula works for any quadratic equation of the form \( ax^2 + bx + c = 0 \). Using this method helps find where the projectile lands or specific heights.

For example, to find where the projectile hits the ground, we set the height equal to zero: \( \frac{-32 x^2}{100^2} + x = 0 \). Factoring out \( x \) results in
\( x \left(-\frac{32 x}{100^2} + 1\right) = 0 \). Therefore, \( x = 0 \) or \( x = 312.5 \) feet.

To find when the height is 50 feet, we solve \( 50 = \frac{-32 x^2}{100^2} + x \) with the quadratic formula. Plugging in \( a = \frac{-32}{10000}, b = 1, \) and \( c = -50 \), we get two solutions:
\( x = 93.75 \) or \( 218.75 \) feet from the origin. This provides an understanding of distances related to given heights on the projectile's path.

Graphing Functions

Graphing functions helps visually interpret the behavior of the quadratic equation describing our projectile's path. For the height function \( h(x)=\frac{-32 x^{2}}{10000}+x \), graphing it reveals a downward-opening parabola due to the negative coefficient of \( x^2 \). By graphing, we can observe the following:

• The vertex, which appears as the peak of the parabola, located at \( (156.25, 78.125) \).
• The roots or x-intercepts, where the function crosses the x-axis, indicating where the projectile hits the ground at \( x = 0 \) and \( x = 312.5 \).
• The general shape and symmetry of the projectile's path around the vertex, implying equal distances on either side for similar heights.

Using graphing utilities assists in verifying these points and understanding the quadratic nature of projectile motion. Graphing tools can help visualize, adjust scales, and confirm results obtained analytically.