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 \(25^{\circ}\) to the horizontal with an initial speed of 500 meters per second.

Short answer

Range \(R \approx 19,558.62\) meters. Maximum height \(H \approx 1,229.64\) meters.

Step-by-step solution

- Identify Known Values

To start, identify the given values: \( \theta = 25^{\circ} \) (launch angle) and \( v_0 = 500 \) meters per second (initial speed).

- Calculate the Time of Flight

Use the formula for the time of flight: \[ T = \frac{2 v_0 \sin(\theta)}{g} \]where \(g\) is the acceleration due to gravity (\(9.8\) m/s²). Substituting the known values, we get:\[ T = \frac{2 \times 500 \times \sin(25^{\circ})}{9.8} \approx 43.42 \text{ seconds} \]

- Calculate the Range

The range \(R\) can be found using the formula:\[ R = v_0 \cos(\theta) \times T \]Substituting the known values and the previously found time of flight, we get:\[ R = 500 \times \cos(25^{\circ}) \times 43.42 \approx 19,558.62 \text{ meters} \]

- Calculate the Maximum Height

Use the formula for maximum height \(H\):\[ H = \frac{(v_0 \sin(\theta))^2}{2g} \]Substituting the known values, we get:\[ H = \frac{(500 \times \sin(25^{\circ}))^2}{2 \times 9.8} \approx 1,229.64 \text{ meters} \]

Key concepts

Range of a Projectile

The range of a projectile, denoted as \( R \), is the total horizontal distance it travels while in motion. It is influenced by both the initial speed and the angle of launch. To calculate the range, we use the formula:

\[ R = v_0 \times \text{cos}(θ) \times T \]

Here:

• \( v_0 \) is the initial speed of the projectile
• \( θ \) is the launch angle
• \( T \) is the time of flight

Substituting the values for the angle (25 degrees) and the speed (500 m/s), and using the previously calculated time of flight (43.42 seconds), we get the range as:

\[ R = 500 \times \text{cos}(25^\text{circ}) \times 43.42 \ \ R \approx 19,558.62 \text{ meters} \]

This means that the projectile will travel approximately 19,558.62 meters horizontally before hitting the ground.

Maximum Height of a Projectile

The maximum height, denoted as \( H \), is the highest point the projectile reaches during its flight. To determine this height, we use the following formula:

\[ H = \frac{(v_0 \times \text{sin}(θ))^2}{2g} \]

In this equation:

• \( v_0 \) is the initial speed of the projectile
• \( θ \) is the launch angle
• \( g \) is the acceleration due to gravity (\text{9.8 m/s^2})

Substituting the given values, we calculate the maximum height:

\[ H = \frac{(500 \times \text{sin}(25^\text{circ}))^2}{2 \times 9.8} \ \ H \approx 1,229.64 \text{ meters} \]

Thus, the projectile will reach a peak height of approximately 1,229.64 meters above the ground.

Time of Flight

The time of flight, denoted as \( T \), is the total time the projectile remains in the air. To find the time of flight, we use the formula:

\[ T = \frac{2 \times v_0 \times \text{sin}(θ)}{g} \]

In this equation:

• \( v_0 \) is the initial speed of the projectile
• \( θ \) is the launch angle
• \( g \) is the acceleration due to gravity (9.8 m/s²)

Substituting the given values for initial speed and launch angle, we get:

\[ T = \frac{2 \times 500 \times \text{sin}(25^\text{circ})}{9.8} \ \ T \approx 43.42 \text{ seconds} \]

This means the projectile will be in the air for approximately 43.42 seconds before it lands back on the ground.