Problem 27 - Calculus III

Question

A projectile is fired at a height of 1.5 m above the ground with an initial velocity of 100 m/sec and at an angle of 30° above the horizontal. Use this information to answer the following questions: Determine the maximum height of the projectile.

Step-by-Step Solution

Answer

The maximum height of the projectile is 129.5 m above the ground.

Step 1: Decompose the Initial Velocity

First, calculate the horizontal and vertical components of the initial velocity. Use trigonometry: \[ v_{0x} = v_0 \cos(\theta) \] and \[ v_{0y} = v_0 \sin(\theta) \] where \( v_0 = 100 \text{ m/s} \) and \( \theta = 30° \). Thus, \[ v_{0x} = 100 \cos(30°) = 100 \times \frac{\sqrt{3}}{2} \approx 86.6 \text{ m/s} \] and \[ v_{0y} = 100 \sin(30°) = 100 \times 0.5 = 50 \text{ m/s} \].

Step 2: Calculate the Time to Reach Maximum Height

The time to reach maximum height is when the vertical velocity becomes zero. Use the formula: \[ v_{y} = v_{0y} - gt \] Set \( v_y = 0 \) and solve for \( t \): \[ 0 = 50 - 9.8t \]\[ t = \frac{50}{9.8} \approx 5.1 \text{ seconds} \].

Step 3: Calculate the Maximum Height Above the Launch Point

Use the formula for vertical displacement: \[ h = v_{0y}t - \frac{1}{2}gt^2 \] Substitute \( v_{0y} = 50 \text{ m/s} \), \( g = 9.8 \text{ m/s}^2 \), and \( t = 5.1 \): \[ h = 50 \times 5.1 - \frac{1}{2} \times 9.8 \times (5.1)^2 \] \[ h = 255 - 127 \approx 128 \text{ m} \].

Step 4: Calculate the Maximum Height Above the Ground

Add the initial height from which the projectile was launched to the height above the launch point to find the maximum height above the ground: \[ \text{Maximum height} = 128 \text{ m} + 1.5 \text{ m} = 129.5 \text{ m} \].

Key Concepts

Trigonometric Decomposition

When a projectile is launched, its initial velocity can be broken down into two components: horizontal and vertical. This is done using trigonometry, specifically the sine and cosine functions.
Consider a projectile launched at an angle \( \theta \) with an initial velocity \( v_0 \). The horizontal component \( v_{0x} \) and vertical component \( v_{0y} \) of the velocity are calculated as follows:

• Horizontal Component: \( v_{0x} = v_0 \cos(\theta) \)
• Vertical Component: \( v_{0y} = v_0 \sin(\theta) \)

For example, if \( v_0 = 100 \text{ m/s} \) and \( \theta = 30^\circ \), then the components become \( v_{0x} = 100 \cos(30^\circ) \approx 86.6 \text{ m/s} \) and \( v_{0y} = 100 \sin(30^\circ) = 50 \text{ m/s} \).
These components are critical in solving projectile motion problems because they allow us to analyze the motion separately in vertical and horizontal directions, making computations straightforward.

Maximum Height Calculation

The maximum height of a projectile occurs when its vertical velocity component becomes zero. At this point, the projectile starts descending. The time \( t \) to reach maximum height can be found using the vertical motion equation \( v_{y} = v_{0y} - gt \), where \( v_y \) is the final vertical velocity (which is 0 at maximum height), \( v_{0y} \) is the initial vertical velocity, and \( g \) is the acceleration due to gravity (9.8 m/s²).

• Setting \( v_y = 0 \) gives us: \( t = \frac{v_{0y}}{g} \).
• With \( v_{0y} = 50 \text{ m/s} \), the time to maximum height is \( t \approx 5.1 \text{ seconds} \).

Once the time is determined, the maximum height \( h \) from the launch point can be calculated using the vertical displacement formula: \[ h = v_{0y}t - \frac{1}{2}gt^2 \]Substituting the known values, \( h \approx 128 \text{ meters} \). Finally, to find the maximum height above the ground, add the launch height (1.5 m), so the peak height is \( 129.5 \text{ meters} \).
This step is vital because it accounts for the complete trajectory of the projectile, ensuring accurate results.

Vertical Motion Equations

Vertical motion equations are essential tools for predicting various aspects of a projectile's flight, like maximum height and time of flight. They describe how an object moves in the vertical direction under the influence of gravity.
The primary equation used in these calculations is the vertical displacement formula:\[ h = v_{0y}t - \frac{1}{2}gt^2 \]where:

• \( h \) is the vertical displacement.
• \( v_{0y} \) is the initial vertical velocity.
• \( t \) is the time elapsed.
• \( g \) is the acceleration due to gravity, approximately 9.8 m/s².

This formula helps determine how high the projectile rises above its starting point after a given time. Additionally, the equation \( v_{y} = v_{0y} - gt \) is used to find the vertical velocity at any moment. These equations allow you to analyze the projectile's motion, making predictions about its trajectory and final position with precision.
Understanding these equations is crucial for solving problems involving projectiles, as they build the framework required for accurate and meaningful conclusions about the motion.