Problem 50 - Algebra and Trigonometry with Analytic Geometry

Question

Projectile's range If a projectile is fired from ground level with an initial velocity of \(v \mathrm{ft} / \mathrm{sec}\) and at an angle of \(\theta\) degrees with the horizontal, the range \(R\) of the projectile is given by $$ R=\frac{v^{2}}{16} \sin \theta \cos \theta . $$ If \(v=80 \mathrm{ft} / \mathrm{sec}\), approximate the angles that result in a range of 150 feet.

Step-by-Step Solution

Answer

Angles are approximately 24.3° and 65.7°.

Step 1: Understand the Formula

The given formula is the range of a projectile, which states that \( R = \frac{v^2}{16} \sin \theta \cos \theta \). Here, \( v \) is the initial velocity, \( \theta \) the angle of projection, and \( R \) the range. We need the angle \( \theta \) when \( v = 80 \text{ ft/sec} \) and \( R = 150 \text{ feet} \).

Step 2: Substitute Given Values

Substitute \( v = 80 \) and \( R = 150 \) into the formula: \[ 150 = \frac{80^2}{16} \sin \theta \cos \theta \]Simplify the expression to find \( \theta \).

Step 3: Simplify the Equation

First, calculate \( \frac{80^2}{16} \): \[ \frac{6400}{16} = 400 \]The equation simplifies to:\[ 150 = 400 \sin \theta \cos \theta \]

Step 4: Reformulate the Equation Using Trigonometric Identity

Use the identity \( \sin 2\theta = 2\sin \theta \cos \theta \) to rewrite the equation:\[ \sin \theta \cos \theta = \frac{1}{2} \sin 2\theta \]The equation becomes:\[ 150 = 200 \sin 2\theta \]

Step 5: Solve for \( \sin 2\theta \)

Solve the equation for \( \sin 2\theta \):\[ \sin 2\theta = \frac{150}{200} = 0.75 \]

Step 6: Find the Possible Angles

Find \( 2\theta \) using the inverse sine function:\[ 2\theta = \sin^{-1}(0.75) \approx 48.59^\circ \]Thus, \( \theta \approx 24.295^\circ \).Since the sine function is positive in both the first and second quadrants:\[ 2\theta = 180^\circ - 48.59^\circ = 131.41^\circ \]This gives another \( \theta \approx 65.705^\circ \).

Step 7: Conclusion with Approximate Angles

Thus, the approximate angles that result in a range of 150 feet are approximately \( 24.3^\circ \) and \( 65.7^\circ \).

Key Concepts

Range of a Projectile

The range of a projectile refers to the horizontal distance it travels when launched from a certain height or level. It is an important aspect of projectile motion that helps determine how far a projectile will land from its initial position. The formula to calculate the range, denoted as \( R \), for a projectile launched at an angle \( \theta \) with initial velocity \( v \) is given by: \[ R = \frac{v^2}{16} \sin \theta \cos \theta \]This equation comes into play when analyzing the flight path of a projectile, allowing us to predict how far it can potentially travel based on initial conditions.

• The numerator \( v^2 \) emphasizes the role of velocity in reaching farther distances.
• The trigonometric components \( \sin \theta \cos \theta \) indicate how the angle of projection influences range.

By understanding how these factors interact, students can grasp why a projectile lands where it does.

Trigonometric Identities

In solving problems involving projectile motion, trigonometric identities are invaluable tools. These identities simplify complex trigonometric expressions, making calculations easier. A vital identity used in this context is:\[ \sin 2\theta = 2 \sin \theta \cos \theta \]This particular identity helps simplify the equation for projectile range. In our problem, after substituting and simplifying the given elements, we used this identity to rewrite the expression:\[ 150 = 200 \sin 2\theta \]It transforms the product of \( \sin \theta \) and \( \cos \theta \) into a more manageable form involving \( \sin 2\theta \).

• These transformations allow for solving trigonometric equations more efficiently.
• They are essential for transitioning between different trigonometric forms.

Mastering these identities enriches a student's ability to deal with trigonometric calculations in physics.

Initial Velocity

The initial velocity \( v \) is the speed at which a projectile starts its motion. It is a critical factor that dictates how far and high a projectile will go. In the projectile range formula, initial velocity is squared: \( v^2 \). This shows that changes in velocity significantly impact the range.

• If you double the velocity, the range quadruples, reflecting a squared relationship.
• The higher the initial speed, the farther the projectile can travel, assuming other conditions stay constant.

Understanding how initial velocity affects motion aids students in predicting and manipulating outcomes in practical scenarios like sports or engineering.

Angle of Projection

The angle of projection \( \theta \) is the angle at which a projectile is launched relative to the horizontal plane. It's one of the most crucial determinants of a projectile's range. The range formula involves both \( \sin \theta \) and \( \cos \theta \), which highlights how optimizing this angle can enhance the projectile's distance. When approached in our exercise, we found:\[ \sin 2\theta = 0.75 \]This resulted in possible angles \( \theta \) of approximately \( 24.3^\circ \) and \( 65.7^\circ \).

• The ideal angle often touted for maximum range is 45 degrees, but practical conditions like wind and elevation might adjust this value.
• Different angles will prioritize height over distance and vice versa.

Grasping the concept of the projection angle allows students to make strategic decisions in tasks involving trajectory.