Conservation of Energy in Ball Thrown Upwards: A Practical Example

In this article, we will explore the physics behind the behavior of a ball thrown upwards, specifically showing that it hits the ground with the same speed it was thrown, assuming no air resistance. This example integrates the principles of conservation of energy and kinematics to provide a clear and practical illustration.

Introduction to the Problem

The problem at hand involves a ball that is thrown upwards with an initial speed of 100 km/h. We aim to prove that the ball hits the ground with the same speed after returning from the highest point, taking into account gravity and neglecting air resistance. The acceleration due to gravity is ( g 10 text{ m/s}^2 ).

Initial Speed Conversion

First, we need to convert the initial speed from kilometers per hour (km/h) to meters per second (m/s).

v_0 100 text{km/h} times frac{1000 text{m}}{1 text{km}} times frac{1 text{h}}{3600 text{s}} frac{1000}{36} approx 27.78 text{m/s}

Step-by-Step Analysis of Motion

Step 1: Time to Reach the Highest Point

At the highest point, the final velocity becomes zero. Using the equation ( v v_0 - g t ), we find the time ( t ) to reach the highest point.

0 27.78 - 10 t implies t frac{27.78}{10} approx 2.778 text{s} Step 2: Maximum Height

Using the kinematic equation for displacement ( h v_0 t - frac{1}{2} g t^2 ), we can find the maximum height.

h 27.78 times 2.778 - frac{1}{2} times 10 times 2.778^2

Calculating the two terms:

First term: ( 27.78 times 2.778 approx 77.11 text{m} ) Second term: ( frac{1}{2} times 10 times 2.778^2 approx 38.61 text{m} )

Thus, the maximum height ( h approx 77.11 - 38.61 38.5 text{m} ).

Step 3: Velocity on Impact

When the ball falls back to the ground, its potential energy at the maximum height is converted back into kinetic energy. At the highest point, the potential energy ( PE mgh ) is equal to the kinetic energy ( KE frac{1}{2} mv^2 ).

gh frac{1}{2} v^2 implies v sqrt{2gh}

Substituting ( g 10 text{ m/s}^2 ) and ( h approx 38.5 text{ m} ):

v sqrt{2 times 10 times 38.5} sqrt{770} approx 27.78 text{m/s}

This shows that the speed just before the ball hits the ground is approximately 27.78 m/s, which is the same as the initial speed of 100 km/h. This confirms the principle of conservation of energy in the absence of air resistance.

Conclusion

By analyzing the motion of the ball and applying the principles of conservation of energy and kinematics, we have demonstrated that a ball thrown upwards with an initial speed of 100 km/h hits the ground with the same speed, assuming no air resistance. This example provides a clear understanding of how the conservation of energy principle operates in practical situations.