Question

Find the range \(R\) and maximum height \(H\) of the projectile. Round answers to two decimal places. The projectile is fired at an angle of \(30^{\circ}\) to the horizontal with an initial speed of 150 meters per second.

Short answer

Range is 1989.17 meters, and maximum height is 286.99 meters.

Step-by-step solution

- Determine Initial Velocity Components

First, find the horizontal and vertical components of the initial velocity. Use the given angle ( 30^{\circ}) and initial speed (150 m/s).Horizontal Component: $$\text{{v}}_{{0x}}=\text{{v}}_0 \cos(\theta) = 150 \cos(30^{\circ}) = 150 \times 0.866 = 129.9 \text{ m/s}.$$Vertical Component: $$\text{{v}}_{{0y}} = \text{{v}}_0 \sin(\theta) = 150 \sin(30^{\circ}) = 150 \times 0.5 = 75 \text{ m/s}.$$

- Find the Time of Flight

The time of flight is the total time the projectile will be in the air. From vertical motion equations,$$t = \frac{{2 \text{{v}}_{{0y}}}}{{g}} = \frac{{2 \times 75}}{{9.8}} \approx 15.31 \text{ seconds}.$$

- Calculate the Range

The range (R) is the horizontal distance traveled by the projectile.$$R = \text{{v}}_{{0x}} t = 129.9 \times 15.31 \approx 1989.17 \text{ meters}.$$

- Calculate Maximum Height

The maximum height (H) is achieved when the vertical component of the velocity is zero:$$H = \frac{{\text{v}_{{0y}}^2}}{2g} = \frac{{75^2}}{2 \times 9.8} \approx 286.99 \text{ meters}.$$

Key concepts

range of projectile

In projectile motion, the range is the horizontal distance the projectile covers before it lands back on the ground. It's important because it tells us how far the object travels.
The formula to calculate the range (\text{R}) of a projectile is:
\[R = v_{0x} t\]
Where

• \(v_{0x}\) is the initial horizontal velocity
• \(t\) is the time of flight

To find \(v_{0x}\), we use:
\[v_{0x} = v_0 \, cos(\theta)\]
Given:
\(v_0 = 150 \ ms^{-1}, \theta = 30^{\text{degrees}}\), we get:
\[v_{0x} = 150 \, cos(30^{\text{degrees}}) = 150 \, \times 0.866 = 129.9 \, ms^{-1}\]
Next, we use the formula for time of flight (discussed in its section) to find \(t\), and then use it to compute the range.
Range calculation:
\[R \approx 129.9 \, \times 15.31 \approx 1989.17 \text{ meters}\].

maximum height of projectile

The maximum height (\text{H}) of a projectile is the highest point it reaches during its motion. It's a useful measure when analyzing the vertical aspect of projectile motion.
The formula to find maximum height is:
\[H = \frac{v_{0y}^2}{2g}\]
Where

• \(v_{0y}\) is the initial vertical velocity
• \(g\) is the acceleration due to gravity

To find \(v_{0y}\), we use:
\[v_{0y} = v_0 \, sin(\theta)\]
Given:
\(v_0 = 150 \ ms^{-1}, \theta = 30^{\text{degrees}}\), we get:
\[v_{0y} = 150 \, sin(30^{\text{degrees}}) = 150 \, \times 0.5 = 75 \, ms^{-1}\]
Then we use this velocity to compute the maximum height:
\[H = \frac{75^2}{2 \times 9.8} \approx 286.99 \text{ meters}\]
This shows that the projectile will reach 286.99 meters above the initial launch point.

initial velocity components

When dealing with projectile motion, the initial velocity is broken down into two components: horizontal (\text{v_{0x}}) and vertical (\text{v_{0y}}). This is crucial because it helps us analyze the motion separately in horizontal and vertical directions.
The formulas are:

• Horizontal: \(v_{0x} = v_0 \, cos(\theta)\)
• Vertical: \(v_{0y} = v_0 \, sin(\theta)\)

Given:
\(v_0 = 150 \ ms^{-1}\), and \(\theta = 30^{\text{degrees}}\), we calculate:

• Horizontal Component: \(v_{0x} = 150 \, cos(30^{\text{degrees}}) = 129.9 \ ms^{-1}\)
• Vertical Component: \(v_{0y} = 150 \, sin(30^{\text{degrees}}) = 75 \ ms^{-1}\)

Understanding these components allows us to use them in further calculations for other properties of projectile motion like range and maximum height.

time of flight

The time of flight (\text{t}) of a projectile is the total time it remains in the air from launch to landing.
To find the time of flight, we focus on the vertical motion using the formula:
\[t = \frac{2v_{0y}}{g}\]
Where

• \(v_{0y}\) is the initial vertical velocity
• \(g\) is the acceleration due to gravity

Given:
\(v_{0y} = 75 \ ms^{-1}, g = 9.8 \ ms^{-2}\), we calculate:
\[t \approx \frac{2 \times 75}{9.8} \approx 15.31 \, \text{seconds}\]
This value indicates that the projectile will be in the air for approximately 15.31 seconds before hitting the ground.