A ball is thrown straight up and passes a window in a multi-story building. The window is2.0 m tall. A person in the room looking out of the window sees the ball move past but itdisappears for a brief interval of time and then it reappears as it falls back down past thewindow. The person sees the ball for a total of 0.80 sec (0.40 sec going up and 0.40 secgoing down). Find the height above the top of the window that the ball rises to before itstarts to fall back down.

To determine how high the ball rises above the top of the window, we can break this problem down into a few logical steps using the principles of kinematics. The key here is to analyze the motion of the ball as it travels upward and then downward, focusing on the time it spends in view of the window.

Understanding the Motion of the Ball

When the ball is thrown straight up, it will decelerate due to gravity until it reaches its highest point, at which point its velocity will be zero before it starts to fall back down. The total time the ball is visible to the observer is 0.80 seconds, with 0.40 seconds for the upward journey and 0.40 seconds for the downward journey.

Using Kinematic Equations

We can use the following kinematic equation to find the maximum height the ball reaches:

• v = u + at
• s = ut + (1/2)at²

Where:

• v = final velocity (0 m/s at the peak)
• u = initial velocity (unknown)
• a = acceleration (−9.81 m/s², due to gravity)
• t = time (0.40 seconds for the upward motion)
• s = displacement (height above the window)

Calculating Initial Velocity

First, we can find the initial velocity (u) using the first equation:

0 = u - 9.81 * 0.40

Rearranging gives:

u = 9.81 * 0.40 = 3.924 m/s

Finding the Maximum Height

Now that we have the initial velocity, we can find the maximum height (s) using the second equation:

s = (3.924 * 0.40) + (1/2)(-9.81)(0.40)²

Calculating each term:

s = (3.924 * 0.40) + (1/2)(-9.81)(0.16)

s = 1.5696 - 0.7848

s = 0.7848 m

Final Height Above the Window

The ball rises approximately 0.785 meters above the top of the window before it starts to fall back down. Therefore, if the window is 2.0 meters tall, the ball will be at a height of:

Height above the ground = height of the window + height above the window

Height above the ground = 2.0 m + 0.785 m = 2.785 m

In summary, the ball reaches a maximum height of about 0.785 meters above the top of the window before descending back down. This analysis illustrates the principles of motion under gravity and the application of kinematic equations to solve real-world problems.