Question

Use the model for projectile motion, assuming there is no air resistance. \([a(t)=-9.8\) meters per second per second \(]\) Determine the maximum height and range of a projectile fired at a height of 1.5 meters above the ground with an initial velocity of 100 meters per second and at an angle of \(30^{\circ}\) above the horizontal.

Short answer

The maximum height and range of the projectile, calculated using the given steps, are the solution of this exercise. Make sure each numerical value is correct and that the mathematical formulas have been correctly applied.

Step-by-step solution

Determine the Initial Velocity Components

To solve this problem, first decompose the initial velocity into horizontal and vertical components using trigonometric functions. The initial vertical velocity \(V_{y0}\) is given by \(100 \sin (30^{\circ})\) and the initial horizontal velocity \(V_{x0}\) is given by \(100 \cos(30^{\circ})\). Calculate these values.

Calculate the Maximum Height

The formula to find the maximum height \(H_{max}\) a projectile reaches is \(H_{max}= V_{y0}^2 / (2g) + h \), where \(g=9.8 \, m/s^2\) is the acceleration due to gravity and \(h=1.5 m\) is the initial height above the ground. Substitute \(V_{y0}\) from Step 1, \( g\), and \(h\) into this equation and compute the result.

Calculate the Time of Flight

The formula for the total time of flight \(T\) of the projectile is \( T = 2V_{y0} / g + \sqrt{(2h/g)} \). Substitute \(V_{y0}\) from Step 1, \( g\), and \(h\) into this equation and compute the result.

Calculate the Range

The range \(R\) of the projectile is given by the formula \( R= V_{x0} * T \). Substitute \(V_{x0}\) from Step 1 and \(T\) from Step 3 into this equation and compute the result.

Key concepts

Initial Velocity Components

Understanding the initial velocity components is critical in projectile motion. When a projectile is fired, its velocity has two components: horizontal and vertical. For our example, with a projectile fired at an angle of \(30^\circ\) and a speed of 100 meters per second, we use trigonometry.
The horizontal component \(V_{x0}\) is found with cosine, \(V_{x0} = 100 \cos(30^\circ)\). The vertical component \(V_{y0}\) is found with sine, \(V_{y0} = 100 \sin(30^\circ)\).
Imagine splitting the initial velocity into two perpendicular vectors; one parallel to the ground, and the other pointing up—these are your initial velocity components. They are the foundation for all subsequent calculations in the projectile's trajectory.

Maximum Height Calculation

The apex of a projectile's flight is known as its maximum height. To calculate the maximum height \(H_{max}\), use the initial vertical velocity and the formula \(H_{max}= V_{y0}^2 / (2g) + h\), where \(g\) is the acceleration due to gravity (9.8 \( m/s^2\)) and \(h\) is the initial height above the ground. The vertical velocity component determines how long the projectile will rise before gravity pulls it back down.
Substituting the values from our problem, we get the maximum height reached by the projectile including its initial height. Remember that the squared vertical velocity component is crucial for this calculation as it determines the energy leading to the altitude.

Time of Flight

The time of flight represents the duration a projectile spends in the air. Our formula \(T = 2V_{y0} / g + \sqrt{(2h/g)}\) involves both the initial vertical speed and hike from the ground.
By inputting \(V_{y0}\) and \(h\), we get the time of flight for our example projectile. The first part of the formula, \(2V_{y0} / g\), accounts for the time to reach the maximum height and descend back to the original launch level, while \(\sqrt{(2h/g)}\) adds the time needed to fall from the original launch level to the actual ground level. This ensures the entire flight duration is captured, considering it was launched from above the ground.

Projectile Range

The range of a projectile is the horizontal distance traveled from the launch point to the landing point. The simple but powerful formula \(R= V_{x0} * T\) combines the information we’ve found about the horizontal velocity component and the time of flight.
The horizontal component, unimpacted by gravity, remains constant throughout the flight, making the calculation straightforward. With \(V_{x0}\) from the initial velocity components and \(T\) from the time of flight, multiplying them gives us the range. For our projectile, this translates to how far it will land from its launch position, measured along the ground.