Problem 14 - Advanced Engineering Mathematics

Question

A small projectile is launched from ground level with an initial speed of \(98 \mathrm{~m} / \mathrm{s}\). Find the possible angles of elevation so that its range is \(490 \mathrm{~m}\).

Step-by-Step Solution

Answer

The angles of elevation are 15° and 75°.

Step 1: Understanding the Problem

The goal is to find the angles of elevation for which a projectile, launched with initial speed of 98 m/s, lands at a distance of 490 m from the launch point. We need to use the projectile motion equations to solve this problem.

Step 2: Use the Range Formula

The range of a projectile, launched at angle \( \theta \) with initial speed \( v_0 \), is given by:\[R = \frac{v_0^2}{g} \sin(2\theta)\]where \( g \approx 9.8 \text{ m/s}^2 \) is the acceleration due to gravity. We are given \( R = 490 \text{ m} \) and \( v_0 = 98 \text{ m/s} \). Substitute these values into the formula.

Step 3: Substitute Known Values

Substitute \( R = 490 \text{ m} \), \( v_0 = 98 \text{ m/s} \), and \( g = 9.8 \text{ m/s}^2 \) into the range formula:\[490 = \frac{98^2}{9.8} \sin(2\theta) \]This simplifies to:\[490 = \frac{9604}{9.8} \sin(2\theta)\]

Step 4: Simplify and Solve for \( \sin(2\theta) \)

Calculate \( \frac{9604}{9.8} \) which results in \( 980 \). Thus, the equation becomes:\[490 = 980 \sin(2\theta)\]Divide both sides by 980 to find \( \sin(2\theta) \):\[\sin(2\theta) = \frac{490}{980} = \frac{1}{2}\]

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

The equation \( \sin(2\theta) = \frac{1}{2} \) has solutions where \( 2\theta \) equals \( 30^\circ \) and \( 150^\circ \) in the range \( 0^\circ \) to \( 180^\circ \).

Step 6: Find \( \theta \)

Divide the solutions for \( 2\theta \) by 2 to find \( \theta \):\[\theta = 15^\circ \quad \text{and} \quad \theta = 75^\circ\] These are the possible launch angles.

Key Concepts

Range Equation

The range equation plays a crucial role in understanding projectile motion. This formula gives us the horizontal distance a projectile travels when it is launched from the ground. The equation is given by:\[R = \frac{v_0^2}{g} \sin(2\theta)\]where:

• The range \(R\) is the horizontal distance covered.
• \(v_0\) is the initial speed of the projectile.
• \(\theta\) is the angle of elevation at which the projectile is launched.
• \(g\) is the acceleration due to gravity, approximately \(9.8 \text{ m/s}^2\).

To use this equation, it is essential to substitute the known values and solve for the unknown variables. In our exercise, we made use of the initial speed \(v_0 = 98 \text{ m/s}\) and the sought-after range \(R = 490 \text{ m}\) to discover the possible angles of elevation.

Angle of Elevation

The angle of elevation, \(\theta\), is the angle at which an object is launched with respect to the horizontal ground level. Adjusting this angle directly affects the range the projectile will achieve. It's important to note that two different angles often produce the same range.
For the exercise, solving the range equation allowed us to find two possible angles of elevation: \(15^\circ\) and \(75^\circ\). These angles satisfy the equation \(\sin(2\theta) = \frac{1}{2}\). By using basic trigonometric solutions, we find that:

• If \(2\theta = 30^\circ\) or \(2\theta = 150^\circ\), then the possible angles are \(\theta = 15^\circ\) or \(\theta = 75^\circ\).

Both angles are valid solutions for the given projectile range.

Initial Speed

The initial speed, represented by \(v_0\), is the speed at which the projectile is launched. It’s one of the fundamental components in determining not just how far, but also how high the projectile will go.
In the context of the exercise, the projectile has an initial speed of \(98 \text{ m/s}\). This high initial speed allows the projectile to achieve a significant range, given the right angle of elevation. As we can see from the range equation, altering the initial speed while holding the angle constant would result in a proportional change in the range.
Understanding the relationship between the initial speed and other factors helps in anticipating the behavior of projectiles in motion.

Acceleration due to Gravity

Gravity is a constant force that affects all objects near the Earth's surface. In physics, we usually denote it by \(g\) and equate it to about \(9.8 \text{ m/s}^2\).
This acceleration is crucial in projectile motion as it influences how quickly the projectile begins descending after reaching its peak height.

• In the range equation, \(g\) aids in determining the vertical component of the initial velocity needed to achieve a certain range.
• It's the factor that causes the characteristic parabolic trajectory of projectile motion.

For the exercise, using \(g = 9.8 \text{ m/s}^2\) allowed us to accurately calculate the potential angles of elevation needed to achieve the desired range of \(490 \text{ m}\).
Understanding gravity in these scenarios helps predict and analyze projectile paths effectively.