Projectile Physics: Calculating Maximum Height Attained by a Ball

In physics, understanding projectile motion is crucial for various applications in sports, engineering, and ballistics. This article will delve into the factors and steps involved in calculating the maximum height attained by a projected ball. We will explore the physics behind it and provide detailed calculations for different scenarios.

Understanding Projectile Motion

Projectile motion occurs when a projectile is launched into the air with an initial velocity and is subject to the acceleration due to gravity. The motion can be broken down into two components: horizontal and vertical, with the vertical component determining the height attained by the projectile.

Calculating Maximum Height: Methodology

The formula to find the maximum height (H) of a projectile launched at an initial velocity v_0 at an angle θ with respect to the horizontal is given by:

H frac{v_0 sin^2 theta}{2g}

Where:

v_0 is the initial velocity, θ is the launch angle, g is the acceleration due to gravity, typically 9.81 m/s2.

Example 1: Velocity 10 m/s, Angle 30°

Let's consider a ball projected with an initial velocity of 10 m/s at an angle of 30° with the horizontal surface.

Calculate the vertical component of the initial velocity:

V_{0y} v_0 sin theta 10 sin 30° 10 times 0.5 5 text{ m/s}

Substitute into the maximum height formula:

H frac{5^2}{2 times 9.81} frac{25}{19.62} approx 1.276 text{ m}

Therefore, the maximum height attained by the projectile is approximately 1.28 meters.

Example 2: Velocity 28 m/s, Angle 30°, Gravity 10 m/s2

Now, let's calculate for a ball projected with a velocity of 28 m/s at an angle of 30° with a gravitational acceleration of 10 m/s2.

Step 1: V_{0y} 28 sin 30° 28 times 0.5 14 text{ m/s}

Step 2: H frac{14^2}{2 times 10} frac{196}{20} 9.8 text{ m}

The maximum height attained by the ball is 9.8 meters.

Example 3: Velocity 10 m/s, Angle 45°

For a ball projected with an initial velocity of 10 m/s at an angle of 45°, the vertical component of the velocity can be calculated as:

V_{0y} 10 sin 45° 10 times frac{sqrt{2}}{2} 10 times 0.707 7.07 text{ m/s}

Substituting into the maximum height formula:

H frac{7.07^2}{2 times 9.81} frac{50}{19.62} approx 2.54 text{ m}

The maximum height attained by the projectile is approximately 2.54 meters.

Additional Calculations

For a projectile with a velocity of 14 m/s at an angle of 30°, the maximum height can be calculated as:

Vertical component of velocity 10 sin 30° 5 meters/s

Using the formula v^2 u^2 - 2gh, where v is final velocity (0), u is initial velocity, and g is acceleration due to gravity:

0 5^2 - 2 times 9.81 times h

25 19.62 h

H frac{25}{19.62} approx 1.274 text{ m}

The maximum height attained is approximately 1.274 meters.

Similarly, for a ball with a velocity of 10 m/s at an angle of 30°, the calculation proceeds as:

Vertical component of velocity 10 sin 30° 5 meters/s

0 5^2 - 2 times 10 times h

25 20 h

H frac{25}{20} 1.25 text{ m}

The maximum height attained is 1.25 meters.

Conclusion

Understanding and calculating the maximum height attained by a projectile is a crucial aspect of projectile motion. By breaking down the problem into its vertical and horizontal components and applying the appropriate formulas, we can determine the height of the projectile at its peak. These calculations are fundamental in various fields and have real-world applications in sports, architecture, and engineering.