Q. 12 A projectile is fired at an incl... [FREE SOLUTION]

91影视

Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 3: Q. 12 (page 165)

A projectile is fired at an inclination of $45 掳$ to the horizontal, with a muzzle velocity of $100 f t / s$ . 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.
Part (a): At what horizontal distance from the firing point is the height of the projectile a maximum?
Part (b): Find the maximum height of the projectile.
Part (c): At what horizontal distance from the firing point will the projectile strike the ground?
Part (d) Using a graphing utility, graph the function h, $0 鈮� x 鈮� 350$ .
Part (e): Use a graphing utility to verify the results obtained in parts (b) and (c).
Part (f): When the height of the projectile is $50 f t$ above the ground, how far has it traveled horizontally?

Short Answer

Expert verified

Part (a): The height of the projectile is a maximum at a horizontal distance of $156.25 f t$ from the firing point.

Part (b): The maximum height of the projectile is $78.125 f t$ .

Part (c): The projectile will strike the ground at a horizontal distance of localid="1647366478966" $312.5 f t$ from the firing point.

Part (d): The required function is given below,

Part (e): The maximum of the function is reached at the point $(156, 78)$ , while the projection will touch the ground for $x = 313$ .

Part (f): The projectile has traveled $62.5 f t$ and $250 f t$ horizontally, when the height is $50 f t$ above the ground.

Step by step solution

Part (a) Step 1. Given information.

Consider the given question,

Function for the height of the projectile is given below,

$h (x) = \frac{- 32 x^{2}}{{(100)}^{2}} + x$

Compare the given function with the standard quadratic equation,

$a = - 0.0032, b = 1, c = 0$

As $a < 0$ , the vertex is the highest point on the parabola that represents the function. The height of the projectile is thus maximum when the value of $x = - \frac{b}{2 a}$ .

Substitute the values in the equation x,

$x = - \frac{1}{2 (- 0.0032)} x = 156.25$

Therefore, the height of the projectile is a maximum at a horizontal distance of $156.25 f t$ from the firing point.

Part (b) Step 1. Find the maximum height.

Substitute $x = 156.25$ in the given function,

$h (156.25) = - \frac{32 {(156.25)}^{2}}{{(100)}^{2}} + 156.25 = 78.125$

Therefore, the maximum height of the projectile is $78.125 f t$ .

Part (c) Step 1. Find the distance of the projectile from the firing point that will strike the ground.

When the projectile strikes the water, the height will be $0$ . Then,

role="math" localid="1647363934493" $0 = \frac{- 32 x^{2}}{{(100)}^{2}} + x x (32 x - 10000) = 0 x = 0, 312.5$

As the distance cannot be $0$ , then the only possible value for x is $312.5$ .

Therefore, the projectile will strike the ground at a horizontal distance of $312.5 f t$ from the firing point.

Part (d) Plot the function.

Plot the function for the interval $0 鈮� x 鈮� 350$ ,

Part (e): Step 1. Plot the function of part (b) and (c).

Consider the given question,

From the graph,

We can see how the maximum of the function is reached at the point $(156, 78)$ , while the projection will touch the ground for $x = 313$ .

Part (f) Step 1. Find the distance when hx=50.

We have to solve the quadratic equation $0.0032 x^{2} - x + 50 = 0$ , use the quadratic formula,

$x = \frac{1 卤 \sqrt{1 - 0.64}}{0.0064} x = 250, 62.5$

Since the path of a projectile is a parabola, it is possible for the ball to reach a height of $50 f t$ twice. The second time the projectile reaches a height of $50 f t$ is after crossing the maximum height.

Therefore, the projectile has traveled $62.5 f t, 250 f t$ horizontally, when the height is $50 f t$ above the ground.