Problem 30 - 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: A projectile is fired from ground level at an angle of \(8^{\circ}\) with the horizontal. The projectile is to have a range of \(50 \mathrm{~m}\). Find the minimum velocity necessary to achieve this range.

Step-by-Step Solution

Answer

The minimum velocity is 42.14 m/s.

Step 1: Understand the Problem

We need to determine the minimum velocity necessary for a projectile launched from the ground at an angle of 8° to achieve a range of 50 meters.

Step 2: Recall the Range Formula

The formula for the range of a projectile is given by \( R = \frac{v^2 \sin(2\theta)}{g} \), where \( R \) is the range, \( v \) is the initial velocity, \( \theta \) is the launch angle, and \( g \) is the acceleration due to gravity \( (9.81 \, \text{m/s}^2) \).

Step 3: Substitute Known Values

Substitute the known values into the range formula: \( R = 50 \, \text{m} \), \( \theta = 8^{\circ} \), and \( g = 9.81 \, \text{m/s}^2 \). The equation becomes \( 50 = \frac{v^2 \sin(16^{\circ})}{9.81} \).

Step 4: Solve for Initial Velocity

Rearrange the formula to solve for \( v \): \( v^2 = \frac{50 \times 9.81}{\sin(16^{\circ})} \). Calculate \( \sin(16^{\circ}) \approx 0.276 \), and then calculate \( v^2 \approx \frac{490.5}{0.276} \approx 1777.17 \). Therefore, \( v \approx \sqrt{1777.17} \approx 42.14 \, \text{m/s} \).

Step 5: Conclusion

The minimum initial velocity necessary to achieve a range of 50 meters at an angle of 8° is approximately 42.14 meters per second.

Key Concepts

Understanding the Range Formula

When discussing projectile motion, one important aspect is the range formula. It helps us determine how far a projectile will travel horizontally if launched at a certain angle and velocity. The formula is expressed as:\[R = \frac{v^2 \sin(2\theta)}{g}\]Let's break this down. The term \(R\) represents the range, or the horizontal distance traveled by the projectile. The symbol \(v\) stands for the initial velocity with which the projectile is fired. The angle \(\theta\) to the horizontal determines the direction, while \(g\) denotes the acceleration due to gravity.Knowing this formula isn't enough. We need to remember that conditions like air resistance can impact actual results in real-life scenarios. However, for many physics problems, we assume ideal conditions where these effects are negligible.

Defining Initial Velocity

Initial velocity is a crucial factor in projectile motion. It is the speed at which the projectile is launched. In our example, we used the range formula to find the minimum initial velocity required for a projectile to travel a specified distance. The importance of initial velocity cannot be understated:

• It directly influences the range the projectile can achieve.
• Higher initial velocities can lead to longer ranges, assuming all other factors remain equal.

In computational terms, initial velocity is often the unknown variable we solve for using mathematical rearrangements, as shown in the given exercise. It was calculated by rearranging the provided formula to isolate this variable.

Setting the Launch Angle

The launch angle, represented as \(\theta\), greatly influences the path of a projectile. This angle is measured from the ground to the direction of the launch.In physics, the angle affects the trajectory's height and distance:

• Angles close to 45° generally maximize range under ideal conditions.
• Lower angles yield lower trajectories but can still achieve considerable distance if coupled with high initial velocity.
• In the exercise, an angle of 8° was used, which required a specific velocity to meet the desired range.

Understanding the link between launch angle and other variables can help in formulating strategies to reach a certain range or height.

The Role of Acceleration Due to Gravity

Gravity is a non-negotiable force impacting all projectiles. In physics calculations, it is usually taken as \(g = 9.81 \text{ m/s}^2\).This value plays a critical role:

• It acts downward on the projectile, affecting its vertical motion.
• The acceleration due to gravity impacts both the maximum height achieved and the time of flight.

For any given launch angle and initial velocity, gravity will always act to bring the projectile back to the ground. Understanding how \(g\) interacts with other factors is essential for accurately predicting projectile behavior.