Projectile motion calculations

The Physics of Projectile Motion: Equations and Calculations

Projectile motion is a form of motion experienced by an object that is thrown near the Earth’s surface and moves along a curved path under the influence of gravity alone. Its study is crucial in fields ranging from sports science to engineering and is usually an introductory topic in physics courses. This article aims to shed light on the basic equations that describe projectile motion and how these equations can be utilized in various calculations.

Basic Assumptions

Before diving into the calculations, let’s establish the basic assumptions for projectile motion:

1. Air resistance is negligible.

2. Acceleration due to gravity (\(g\)) is constant.

3. The Earth is flat over short distances.

4. The motion is two-dimensional.

Key Equations

Horizontal Motion

The horizontal distance (\(x\)) is calculated by:

\[

x = v_{0x} t

\]

Where:

– \(x\) is the horizontal distance.

– \(v_{0x}\) is the initial horizontal velocity.

– \(t\) is the time.

Vertical Motion

The vertical distance (\(y\)) can be calculated using the following equation:

\[

y = v_{0y}t – \frac{1}{2}gt^2

\]

Where:

– \(y\) is the vertical distance.

– \(v_{0y}\) is the initial vertical velocity.

– \(g\) is the acceleration due to gravity.

– \(t\) is the time.

Time of Flight

The total time (\(T\)) the projectile is in motion can be expressed as:

\[

T = \frac{2v_{0y}}{g}

\]

Maximum Height

The maximum height (\(H\)) reached by the projectile is:

\[

H = \frac{v_{0y}^2}{2g}

\]

Range

The total horizontal distance (\(R\)) covered by the projectile is:

\[

R = \frac{v_{0}^2 \sin 2\theta}{g}

\]

Where \(v_0\) is the initial velocity and \(\theta\) is the launch angle.

Example Calculations

Example 1: Time of Flight

Given \(v_{0y} = 20 \, \text{m/s}\) and \(g = 9.81 \, \text{m/s}^2\):

\[

T = \frac{2 \times 20}{9.81} \approx 4.08 \, \text{s}

\]

Example 2: Maximum Height

Given \(v_{0y} = 20 \, \text{m/s}\) and \(g = 9.81 \, \text{m/s}^2\):

\[

H = \frac{(20)^2}{2 \times 9.81} \approx 20.4 \, \text{m}

\]

Example 3: Range

Given \(v_0 = 30 \, \text{m/s}\), \(\theta = 45^\circ\), and \(g = 9.81 \, \text{m/s}^2\):

\[

R = \frac{30^2 \sin 90^\circ}{9.81} \approx 91.6 \, \text{m}

\]

Conclusion

Understanding the equations governing projectile motion allows us to make various calculations about an object’s path, time of flight, maximum height, and range. This knowledge is not only academically enriching but also practically useful in various scientific and engineering applications.

**PROBLEMS AND SOLUTIONS**

Conceptual Questions

1. What are the forces acting on a projectile in motion?

Solution: The only force acting on a projectile in motion is gravity.

2. Is the horizontal velocity of a projectile constant?

Solution: Yes, the horizontal velocity of a projectile is constant in the absence of air resistance.

3. What happens to the vertical velocity of a projectile at its peak?

Solution: The vertical velocity of a projectile at its peak is zero.

4. What is the trajectory shape of a projectile launched in the air?

Solution: The trajectory shape of a projectile is parabolic.

5. What factors affect the range of a projectile?

Solution: Factors affecting the range include the initial velocity and the launch angle. Air resistance can also affect the range but is often neglected in basic problems.

Calculation Questions

6. Calculate the time of flight for a projectile launched with an initial vertical velocity of \( 25 \, \text{m/s} \).

Solution:

\[

T = \frac{2 \times 25}{9.81} \approx 5.1 \, \text{s}

\]

7. What is the maximum height reached by a projectile with an initial vertical velocity of \( 25 \, \text{m/s} \)?

Solution:

\[

H = \frac{25^2}{2 \times 9.81} \approx 31.9 \, \text{m}

\]

8. A projectile is launched with an initial velocity of \( 20 \, \text{m/s} \) at an angle of \( 30^\circ \). What is its initial vertical velocity?

Solution:

\[

v_{0y} = 20 \times \sin(30^\circ) = 20 \times 0.5 = 10 \, \text{m/s}

\]

9. Using the same initial conditions as Question 8, what is the initial horizontal velocity?

Solution:

\[

v_{0x} = 20 \times \cos(30^\circ) = 20 \times \frac{\sqrt{3}}{2} \approx 17.3 \, \text{m/s}

\]

10. Calculate the range of a projectile launched with an initial velocity of \( 30 \, \text{m/s} \) at \( 45^\circ \).

Solution:

\[

R = \frac{30^2 \times \sin(90^\circ)}{9.81} \approx 91.7 \, \text{m}

\]

11. What is the time of flight for a projectile launched with an initial velocity of \( 50 \, \text{m/s} \) at \( 60^\circ \)?

Solution:

\[

T = \frac{2 \times 50 \times \sin(60^\circ)}{9.81} \approx 8.8 \, \text{s}

\]

12. What is the maximum height reached by a projectile launched with an initial velocity of \( 40 \, \text{m/s} \) at \( 60^\circ \)?

Solution:

\[

H = \frac{(40 \times \sin(60^\circ))^2}{2 \times 9.81} \approx 94.2 \, \text{m}

\]

13. Calculate the range of a projectile launched with an initial velocity of \( 15 \, \text{m/s} \) at \( 30^\circ \).

Solution:

\[

R = \frac{15^2 \times \sin(60^\circ)}{9.81} \approx 22.4 \, \text{m}

\]

14. What is the time of flight for a projectile launched with an initial velocity of \( 35 \, \text{m/s} \) at \( 70^\circ \)?

Solution:

\[

T = \frac{2 \times 35 \times \sin(70^\circ)}{9.81} \approx 7.2 \, \text{s}

\]

15. A projectile is launched with an initial velocity of \( 20 \, \text{m/s} \) at \( 30^\circ \). What is its vertical velocity after 2 seconds?

Solution:

\[

v_{y} = v_{0y} – gt = 20 \times \sin(30^\circ) – 9.81 \times 2 = 10 – 19.62 = -9.62 \, \text{m/s}

\]