Problem 178 - Calculus Volume 3

Question

A projectile is fired at a height of \(1.5 \mathrm{~m}\) above the ground with an initial velocity of \(100 \mathrm{~m} / \mathrm{sec}\) and at an angle of \(30^{\circ}\) 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 128.90 meters.

Step 1: Determine the Vertical Component of Velocity

The initial velocity is given as \( v_0 = 100 \, \text{m/s} \), and the angle of projection is \( \theta = 30^\circ \). We need to find the vertical component of this velocity using the formula \( v_{0y} = v_0 \sin\theta \). Thus, \( v_{0y} = 100 \sin(30^\circ) = 100 \times 0.5 = 50 \, \text{m/s} \).

Step 2: Use the Kinematic Equation for Maximum Height

The maximum height is reached when the vertical component of the velocity becomes zero. Use the kinematic equation: \( v_y^2 = v_{0y}^2 - 2g(h - h_0) \). Here, \( v_y = 0 \) (at the maximum height), \( v_{0y} = 50 \, \text{m/s} \), \( g = 9.81 \, \text{m/s}^2 \), and \( h_0 = 1.5 \, \text{m} \). Solving for \( h \), we get: \[ 0 = 50^2 - 2 \times 9.81 \times (h - 1.5) \].

Step 3: Solve for Maximum Height

Rearrange the equation from Step 2: \( 50^2 = 2 \times 9.81 \times (h - 1.5) \). Simplify to get \( h - 1.5 = \frac{2500}{19.62} \). Calculate \( h:\) \( h = \frac{2500}{19.62} + 1.5 \). Thus, \( h = 127.40 + 1.5 \), so \( h = 128.90 \). The maximum height of the projectile is \( 128.90 \, \text{m} \).

Key Concepts

Kinematic Equations

Kinematic equations are essential in understanding projectile motion. They relate to the motion of an object under constant acceleration, such as gravity. In projectile motion, kinematic equations help us calculate key aspects such as velocity, time, and displacement.
In our exercise, kinematic equations are used to determine how the height of a projectile changes over time. Specifically, we use the equation:

• \[ v_y^2 = v_{0y}^2 - 2g(h - h_0) \]
• This equation indicates the relationship between the final vertical velocity squared \( v_y^2 \), the initial vertical velocity squared \( v_{0y}^2 \), the acceleration due to gravity \( g \), and the change in height \( h - h_0 \).
  
  Understanding these relationships allows us to find out when the projectile reaches its maximum height, which is when the vertical velocity becomes zero.

Vertical Component of Velocity

The vertical component of velocity is a crucial aspect of any projectile motion problem. It represents the part of an object's velocity that affects its upward and downward movement.
In projectile mechanics, the velocity is often broken down into horizontal and vertical components using trigonometric functions. Given an initial velocity and an angle of projection, we can determine the vertical component of velocity with the formula:

• \[ v_{0y} = v_0 \sin(\theta) \]
• where \( v_0 \) is the initial velocity and \( \theta \) is the angle above the horizontal.
  
  For the specific exercise we are analyzing, the vertical component was calculated to be \( 50 \, \text{m/s} \) with an angle of \( 30^{\circ} \). Having this vertical component is vital because it helps in determining how high the projectile will go before it starts descending again.
  By understanding and calculating this value, we get closer to solving the problem of finding the maximum height.

Maximum Height Calculation

Calculating the maximum height of a projectile involves understanding when its vertical motion halts momentarily.This is the point where the projectile stops rising and starts falling back down.
The maximum height is achieved when the vertical component of the velocity \( v_y \) becomes zero. Using the equation:

• \[ 0 = v_{0y}^2 - 2g(h - h_0) \]
• we can solve for \( h \), the maximum height. In our case:
  
  • Set \( v_y = 0 \) for when the projectile is at its peak.
  • Rearrange the equation to find \( h \).
  • Solve: \( h = \frac{2500}{19.62} + 1.5 \)
    
  
  These calculations led to a maximum height of approximately \( 128.90 \, \text{meters} \). This demonstrates the practical use of mathematics in predicting real-world outcomes, showcasing the power of physics and its equations in solving tangible problems.