Problem 26 - Calculus Early Transcendentals

Question

A gun is fired with angle of elevation \(30^{\circ} .\) What is the muzzle speed if the maximum height of the shell is 500 \(\mathrm{m} ?\)

Step-by-Step Solution

Answer

198 m/s.

Step 1: Understanding the Problem

To find the muzzle speed, we'll use the given maximum height of the projectile. The highest point reached by the projectile corresponds to the time when its vertical velocity component becomes zero.

Step 2: Identify Relevant Formula

The formula to find the maximum height reached by a projectile is: \[ h_{max} = \frac{v_0^2 \sin^2 \theta}{2g} \]where \( v_0 \) is the initial velocity (muzzle speed), \( \theta \) is the angle of projection (30°), and \( g \) is the acceleration due to gravity (9.8 m/s").

Step 3: Substitute Known Values into Formula

We know that the maximum height \( h_{max} = 500 \) meters and \( \theta = 30^{\circ} \). Substituting these values into the formula gives:\[ 500 = \frac{v_0^2 \sin^2 30^{\circ}}{2 \times 9.8} \].

Step 4: Calculate \( \sin 30^{\circ} \)

Recall that \( \sin 30^{\circ} = \frac{1}{2} \).

Step 5: Simplifying the Equation

Substituting \( \sin 30^{\circ} = \frac{1}{2} \) in the equation gives:\[ 500 = \frac{v_0^2 \left(\frac{1}{2}\right)^2}{2 \times 9.8} \]. Simplifying further:\[ 500 = \frac{v_0^2 \frac{1}{4}}{19.6} \].

Step 6: Solve for \( v_0 \)

To find \( v_0 \), solve the equation:\[ 500 = \frac{v_0^2}{78.4} \] Multiply both sides by 78.4:\[ 500 \times 78.4 = v_0^2 \] \[ 39200 = v_0^2 \]Take the square root of both sides:\[ v_0 = \sqrt{39200} \approx 198 \text{ m/s} \].

Key Concepts

Muzzle Speed Calculation

Muzzle speed refers to the initial speed at which a projectile is launched. Calculating this speed is pivotal for understanding projectile motion, especially when firing a projectile at an angle.
To find the muzzle speed based on the maximum height, we use the formula for the height reached by a projectile:

• \[ h_{max} = \frac{v_0^2 \sin^2 \theta}{2g} \]

In this formula:

• \( v_0 \) is the initial velocity or muzzle speed.
• \( \theta \) is the launch angle (in this case, 30 degrees).
• \( g \) is the acceleration due to gravity, approximately \( 9.8 \text{ m/s}^2 \).

Substituting the known values, we isolate \( v_0 \) and solve for it. This involves some algebraic manipulation, starting with setting the given maximum height to 500 meters and replacing \( \sin 30^{\circ} \) with 0.5. Calculating step-by-step, we find fall the muzzle speed to be approximately 198 m/s.
This type of calculation is crucial in physics, where understanding the properties of motion can help predict the path and impact point of a projectile.

Maximum Height of Projectile

The maximum height of a projectile is the peak point in its trajectory, where its vertical velocity component momentarily becomes zero before descending back to the ground.
To determine this height, we analyze the vertical motion of the projectile. At this peak, the kinetic energy in the vertical direction is converted into potential energy.
The relevant formula for maximum height stems from combining initial velocity components with gravitational forces:

• \[ h_{max} = \frac{v_0^2 \sin^2 \theta}{2g} \]

Here, \( \sin^2 \theta \) ensures we focus only on the vertical component of velocity.
This consideration is vital, as the projectile's launch angle heavily influences its vertical dynamics.

• A larger launch angle increases the proportion of vertical motion, potentially raising the maximum height.
• Conversely, a smaller launch angle reduces the vertical component, lowering the peak height.

In our exercise, with a set maximum height of 500 meters, understanding these dynamics allows us to back-calculate the key component of initial speed required to achieve this height.

Trigonometric Functions in Physics

Trigonometric functions play a fundamental role in projectile motion, especially when analyzing the components of motion at angles. Understanding these functions is crucial in deriving predictable outcomes from projectile dynamics.
In physics, we primarily use the sine, cosine, and tangent functions to resolve the initial velocity of a projectile into horizontal and vertical components. For this problem:

• The sine function (\( \sin \)) helps isolate the vertical component.
• The cosine function (\( \cos \)) isolates the horizontal component (though not the primary focus in calculating maximum height).

With the given problem, where a projectile is fired at a 30-degree angle, knowing that \( \sin 30^{\circ} = \frac{1}{2} \) is beneficial. This value allows us to focus on the vertical aspect of motion effectively, translating initial speed predictions into tangible height measurements.
Physics immensely benefits from these functions as they help break down complex movement into manageable calculations, providing insights into range, height, and time for any projectile.