Calculating the Height a Ball Rises Before Passing Through a 2 Meter Window

The motion of a ball passing through a window of a certain height is an intriguing problem in physics. Let us delve into the calculations to understand the height the ball rises before it goes through the window, given the duration of its sight is 1 second and the window height is 2 meters.

Understanding the Problem and Key Concepts

This question revolves around the principles of physics, specifically the kinematics of motion under constant acceleration due to gravity. The key components to consider are the total time the ball is in sight, the height of the window, and the relationships between time, velocity, displacement, and acceleration.

Given Data

Total time the ball is in sight: 1 second Height of the window: 2 meters

Step-by-Step Analysis

To find the height above the window that the ball rises, we can utilize the equations of motion under constant acceleration. Let's break it down step by step.

1. Time Analysis

When the ball goes up, it takes half of the total time to reach its highest point and then takes the same amount of time to come down to the window height. Thus,

(t_{up} frac{t_{total}}{2} frac{1 , text{s}}{2} 0.5 , text{s})

2. Velocity and Acceleration

Using the equation for displacement under uniform acceleration:

(h v_0 t - frac{1}{2} a t^2)

Where:

(v_0) is the initial velocity (a) is the acceleration, which is (-g approx -9.81 , text{m/s}^2), acting downwards (t) is the time

3. Calculating Initial Velocity

At the highest point, the final velocity is 0. Using the first equation of motion:

(v v_0 - g t_{up})

Setting (v 0) at the peak:

(0 v_0 - g t_{up})

Solving for (v_0):

(v_0 g t_{up} 9.81 , text{m/s}^2 times 0.5 , text{s} 4.905 , text{m/s})

4. Calculating Height Above the Window

Now we can find the height above the window using (v_0) in the height equation:

(h_{above} v_0 t_{up} - frac{1}{2} g t_{up}^2)

Substituting the values:

(h_{above} 4.905 , text{m/s} times 0.5 , text{s} - left(frac{1}{2} times 9.81 , text{m/s}^2 times 0.5 , text{s}^2right))

Calculating each term:

(v_0 t_{up} 4.905 times 0.5 2.4525 , text{m}) (frac{1}{2} g t_{up}^2 frac{1}{2} times 9.81 times 0.25 1.22625 , text{m})

Now substituting these values:

(h_{above} 2.4525 , text{m} - 1.22625 , text{m} 1.22625 , text{m})

Conclusion: The height above the window that the ball rises is approximately 1.23 meters.

Alternative Calculation

Alternatively, let (t) be the total time for the ball to fall down, and let the height above the window be (h). The initial velocity is (u 0), and (g 10 , text{m/s}^2). The equation for displacement is:

(s ut frac{1}{2}gt^2)

Given that the total height is (2 - h) and the total time is (t - frac{1}{4}), we get:

(2 - h 0 cdot (t - frac{1}{4}) frac{1}{2} cdot 10 cdot (t - frac{1}{4})^2)

Substituting the values and solving for (h):

(5t^2 5t^2 - frac{5t}{2} frac{27}{16})

(frac{5t}{2} -frac{27}{16})

(t frac{27}{40})

Substituting (t frac{27}{40}) into the height equation:

(h frac{5 cdot (27)^2}{1600} frac{2729}{1600} 2.278 , text{m})

Conclusion: Using this alternative method, the height above the window that the ball rises is approximately 2.278 meters.

Key Takeaways

The height of the window is 2 meters. The total time the ball is in sight is 1 second. The initial velocity of the ball is 4.905 m/s. The height above the window that the ball rises is approximately 1.23 meters using one calculation method and 2.278 meters using another method.

Understanding these calculations helps in comprehending the motion of objects under gravitational force, a fundamental concept in physics and essential for students and professionals alike.