Water Balloon Launch: Understanding Trajectory and Max Height Through Simple Physics

Engage in the fascinating world of projectile motion with the seemingly simple yet profoundly interesting case of a water balloon being catapulted into the air. Through the application of basic physics principles, we can understand the trajectory, velocity, and height of a water balloon launched with a catapult. This article will break down the steps and equations used to calculate the height of a water balloon after a set amount of time, providing you with the knowledge to accurately predict where and when the water balloon will hit its peak and finally land.

Understanding the Trajectory Equation

Regardless of the complexity of the trajectory, the height ( H ) of the water balloon after ( T ) seconds can be determined using a quadratic equation. This equation reflects the parabolic path followed by the balloon in the air:

( H(T) -5T^2 20T 25 )

This equation has a significant physical meaning: the coefficient (-5) represents the acceleration due to gravity (in meters per second squared), (20T) indicates the initial velocity of the water balloon, and (25) is the initial height from which the balloon is launched.

Calculating the Height After 1 Second

Let’s dive into the process of calculating the height of the water balloon after 1 second using the given equation:

Substitute ( T 1 ) into the equation to find the height after 1 second:

( H(1) -5(1)^2 20(1) 25 )

Calculate each term:

Sum the terms:

( H(1) -5 20 25 )

( H(1) 40 ) meters

So, the height of the water balloon after 1 second is 40 meters. This calculation utilizes the variables in the equation to determine the position of the balloon at that exact moment in time.

The Mathematics Behind the Trajectory

The quadratic equation ( H(T) -5T^2 20T 25 ) is a mathematical model that describes the height of the water balloon over time. To fully understand the trajectory, we can also analyze its vertex, where the balloon reaches its maximum height. The vertex form of a parabola is given by:

( T_{text{max}} -frac{b}{2a} )

Here, ( a -5 ) and ( b 20 ). Thus, the time ( T_{text{max}} ) when the balloon reaches its maximum height can be calculated as:

( T_{text{max}} -frac{20}{2(-5)} )

( T_{text{max}} frac{20}{10} )

( T_{text{max}} 2 ) seconds

Substituting ( T 2 ) back into the equation to find the maximum height:

( H(2) -5(2)^2 20(2) 25 )

( H(2) -5 times 4 20 times 2 25 )

( H(2) -20 40 25 )

( H(2) 45 ) meters

This reveals that the maximum height reached by the water balloon is 45 meters, illustrating the peak of its parabolic trajectory.

Conclusion

The trajectory of a water balloon launched by a catapult is a beautiful example of physics in action. By applying basic equations, we can predict its velocity and height at any given time. Using the equation ( H(T) -5T^2 20T 25 ), we calculated the height of the balloon after 1 second, finding it to be 40 meters. Through the understanding of the vertex, we also determined the maximum height to be 45 meters. This insight into projectile motion not only deepens our appreciation for the mathematics behind real-world phenomena but also provides valuable practical knowledge for enthusiasts and professionals.