Problem 53 - Physics for Scientists and Engineers with Modern Physics

Question

A projectile is fired with an initial speed of \(46.6 \mathrm{~m} / \mathrm{s}\) at an angle of \(42.2^{\circ}\) above the horizontal on a long flat firing range. Determine \((a)\) the maximum height reached by the projectile, \((b)\) the total time in the air, \((c)\) the total horizontal distance covered (that is, the range), and \((d)\) the velocity of the projectile \(1.50 \mathrm{~s}\) after firing.

Step-by-Step Solution

Answer

(a) 49.2 m, (b) 6.34 s, (c) 219.4 m, (d) 38.3 m/s.

Step 1: Break Down Initial Velocity Components

Start by identifying the horizontal and vertical components of the initial velocity. Use the formulas: \( V_{0x} = V_0 \cos(\theta) \) and \( V_{0y} = V_0 \sin(\theta) \), where \( V_0 = 46.6 \space \mathrm{m/s} \) and \( \theta = 42.2^\circ \).

Step 2: Calculate Initial Velocity Components

Calculate:\[ V_{0x} = 46.6 \cos(42.2^\circ) \approx 34.6 \mathrm{~m/s} \]\[ V_{0y} = 46.6 \sin(42.2^\circ) \approx 31.1 \mathrm{~m/s} \]

Step 2: Determine the Maximum Height

Use the formula for maximum height: \( h = \frac{V_{0y}^2}{2g} \), where \( g = 9.81 \mathrm{~m/s^2} \) is the acceleration due to gravity.

Step 4: Calculate Maximum Height

Plug in the values:\[ h = \frac{(31.1)^2}{2 \times 9.81} \approx 49.2 \mathrm{~m} \]

Step 3: Determine Total Time in Air

Use the formula for time of flight: \( t = \frac{2V_{0y}}{g} \).

Step 6: Calculate Total Time in Air

Plug in the values:\[ t = \frac{2 \times 31.1}{9.81} \approx 6.34 \mathrm{~s} \]

Step 4: Determine the Range

Calculate the horizontal range using: \( R = V_{0x} \times t \).

Step 8: Calculate the Range

Plug in the values:\[ R = 34.6 \times 6.34 \approx 219.4 \mathrm{~m} \]

Step 5: Determine Velocity After 1.50s

Use kinematic equations to find the velocity components at \( t = 1.5 \mathrm{~s} \). For horizontal velocity, \( V_x = V_{0x} \). For vertical velocity, \( V_y = V_{0y} - g \times t \).

Step 10: Calculate Velocity After 1.50s

Compute:\[ V_x = 34.6 \mathrm{~m/s} \]\[ V_y = 31.1 - 9.81 \times 1.5 \approx 16.3 \mathrm{~m/s} \]The total velocity is:\[ V = \sqrt{V_x^2 + V_y^2} \approx \sqrt{34.6^2 + 16.3^2} \approx 38.3 \mathrm{~m/s} \]

Key Concepts

Kinematics

Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause this motion. Projectile motion, a key concept in kinematics, describes the motion of an object thrown or projected into the air, subject to only the acceleration of gravity.

Key characteristics of projectile motion include the object following a curved path, called a parabola, due to the influence of gravity. The motion can be split into two main components: horizontal and vertical. Each of these components can be analyzed separately using kinematic equations.

For example:

• The horizontal motion is constant because no external force acts in the horizontal direction (assuming air resistance is negligible).
• The vertical motion is affected by gravity, which causes the object to accelerate downward.

Understanding kinematics allows us to predict the future position, velocity, and other important parameters of a moving object.

Velocity Components

To fully understand projectile motion, it's important to break down the initial velocity into two components: horizontal (\( V_{0x} \)) and vertical (\( V_{0y} \)). These components are essential for calculating key characteristics like maximum height and range.

Here is how we determine the velocity components:

• Horizontal component: \( V_{0x} = V_0 \cos(\theta) \), where \( V_0 \) is the initial speed and \( \theta \) is the launch angle.
• Vertical component: \( V_{0y} = V_0 \sin(\theta) \).

In our example, given an initial speed of \(46.6 \space \mathrm{m/s}\) and angle \(42.2^{\circ}\), the components are calculated as follows:

• \( V_{0x} \approx 34.6 \space \mathrm{m/s}\)
• \( V_{0y} \approx 31.1 \space \mathrm{m/s}\)

These components are crucial for determining other features of the projectile's flight.

Maximum Height

The maximum height of a projectile is the highest vertical position in its trajectory. To find this, we focus on the vertical motion component. The formula to calculate the maximum height is:
\[ h = \frac{V_{0y}^2}{2g} \] where \( g = 9.81 \space \mathrm{m/s^2} \) is the acceleration due to gravity.

This formula comes from the understanding that at the maximum height, the vertical component of the velocity becomes zero for a brief moment before the projectile starts descending. Using the initial vertical velocity \( V_{0y} \approx 31.1 \space \mathrm{m/s} \), the calculated maximum height in our example is approximately \( 49.2 \space \mathrm{m} \). This height represents the point where all upward motion ceases, and the projectile starts its downward journey.

Range

The range of a projectile is the total horizontal distance it travels before hitting the ground. Understanding the components of velocity is critical in determining the range. The range can be calculated by analyzing only the horizontal motion, which remains constant.

The formula for range \( R \) is:
\[ R = V_{0x} \times t \] where \( t \) is the total time the projectile is in the air.

By using the values \( V_{0x} \approx 34.6 \space \mathrm{m/s} \) and \( t \approx 6.34 \space \mathrm{s} \), we find that the range is approximately \( 219.4 \space \mathrm{m} \). The range tells us how far the projectile lands from its launch position, assuming the ground is level.

Time of Flight

The time of flight is the total duration from when the projectile is launched until it returns to the same vertical level as the launch point. For symmetrical projectile motion, the time of flight depends primarily on the vertical component of motion.

The time of flight can be calculated using the formula: \[ t = \frac{2V_{0y}}{g} \] where \( V_{0y} \) is the initial vertical velocity, and \( g \) is the acceleration due to gravity.

In our scenario, with \( V_{0y} \approx 31.1 \space \mathrm{m/s} \), the total time in the air is approximately \( 6.34 \space \mathrm{s} \). This concept is crucial in solving problems related to the motion duration, as it helps determine how long the projectile remains airborne.