Problem 173 - Calculus Volume 3

Question

A projectile is shot in the air from ground level with an initial velocity of \(500 \mathrm{~m} / \mathrm{sec}\) at an angle of \(60^{\circ}\) with the horizontal. The graph is shown here: . At what time does the projectile reach maximum height?

Step-by-Step Solution

Answer

The projectile reaches maximum height at approximately 44.2 seconds.

Step 1: Identify Components of Velocity

To analyze projectile motion, we first need to break down the initial velocity of the projectile into its horizontal and vertical components. Given an initial velocity \( v_0 = 500 \, \text{m/s} \) at an angle \( \theta = 60^\circ \), the vertical component of the velocity is \( v_{0y} = v_0 \sin \theta = 500 \sin 60^\circ \).

Step 2: Calculate Vertical Velocity Component

Using trigonometry, we calculate the vertical component: \( v_{0y} = 500 \sin 60^\circ = 500 \times \frac{\sqrt{3}}{2} = 250 \sqrt{3} \approx 433 \text{ m/s}\).

Step 3: Use Kinematics Equation for Maximum Height

The time to reach the maximum height is determined by the equation \( v_y = v_{0y} - gt \), where \( v_y = 0 \) at maximum height. Rearrange to find \( t = \frac{v_{0y}}{g} \).

Step 4: Substitute Values into Equation

Substitute \( v_{0y} = 433 \text{ m/s} \) and \( g = 9.8 \text{ m/s}^2 \) into the equation: \( t = \frac{433}{9.8} \approx 44.2 \text{ seconds} \).

Key Concepts

Kinematics

Kinematics is the study of motion without considering the forces that cause the motion. In projectile motion, kinematics helps us understand how objects move through space and time. With kinematics, we're interested in parameters like velocity, displacement, and acceleration.
Each of these is analyzed in either a horizontal or vertical direction, especially for projectiles moving in two dimensions.

• **Displacement** refers to the change in position of an object. It can be broken into horizontal (x) and vertical (y) components.
• **Velocity** is the rate of change of displacement and has both magnitude and direction. It also has horizontal and vertical components.
• **Acceleration** in projectile motion is usually due to gravity, affecting only the vertical motion. The horizontal motion, if we neglect air resistance, is a constant velocity motion.

Understanding these elements allows us to predict the path of a projectile and calculate key measures like time of flight, range, and maximum height.

Initial Velocity

Initial velocity is crucial in determining a projectile's motion. It describes the speed and direction at which the projectile is launched. In this exercise, the initial velocity is given as 500 m/s at a 60° angle with the horizontal.

• The initial velocity can be split into two components: horizontal and vertical.
• To find these components, we use trigonometry: the horizontal component is \( v_{0x} = v_0 \cos \theta \) and the vertical component is \( v_{0y} = v_0 \sin \theta \).
• For the given problem, the vertical component, which impacts the maximum height, is calculated as \( 250 \sqrt{3} \approx 433 \text{ m/s}\).

These components help determine how far and how high the projectile will travel.

Trigonometry

Trigonometry plays an essential role in analyzing projectile motion, especially when decomposing the initial velocity into its components. By using trigonometric functions like sine and cosine, we can accurately determine horizontal and vertical velocity components.

• The sine function helps find the vertical component: \( v_{0y} = v_0 \sin \theta \).
• The cosine function helps find the horizontal component: \( v_{0x} = v_0 \cos \theta \).
• For angles like 60°, commonly seen in physics problems, it's helpful to remember that \( \sin 60^\circ = \frac{\sqrt{3}}{2} \) and \( \cos 60^\circ = \frac{1}{2} \).

Applying these functions lets us transition from a single vector into separate x and y motion components for more precise calculations.

Maximum Height

The maximum height in projectile motion is the highest point a projectile reaches before descending back to the ground. At this point, the vertical component of the velocity becomes zero.

• We can find the time it takes to reach the maximum height using the formula \( v_y = v_{0y} - gt \).
• At maximum height, \( v_y = 0 \), which allows us to solve for time: \( t = \frac{v_{0y}}{g} \).
• Substituting the values \( v_{0y} = 433 \text{ m/s} \) and \( g = 9.8 \text{ m/s}^2 \), we find \( t \approx 44.2 \text{ seconds} \).

Understanding how to calculate maximum height is essential for solving a range of projectile motion problems.