A ball is dropped from the roof of a building. It hits the ground and it is caught at its original height 5.0 s later. a) What was the speed of the ball just before it hits the ground? b) How tall was the building? c) You are watching from a window \(2.5 \mathrm{~m}\) above the ground. The window opening is \(1.2 \mathrm{~m}\) from the top to the bottom. At what time after the ball was dropped did you first see the ball in the window?

Free fall motion describes the movement of an object under the influence of gravity alone, with no other forces acting upon it. During free fall, an object accelerates downwards due to the force of gravity, which on Earth is approximately \( 9.8 \, \mathrm{m/s^2} \).

This constant acceleration impacts various properties of the falling object, such as its velocity and displacement over time. The two key aspects to note about free fall are that any object experiencing free fall has an initial speed of zero when it begins to drop, and the only acceleration it encounters is due to gravity, which is constant and directed towards the center of the Earth.

Understanding free fall is crucial for analyzing the motion of projectiles or objects dropped from heights, such as in the question about the ball dropped from the top of a building. To calculate the time it takes for the ball to hit the ground, we divide the total time of the round trip by two, since the time up equals the time down in free fall with no air resistance.

Calculating the final velocity of an object in free fall is a common problem in kinematics and physics. To determine this value, we can employ the kinematic equation \(v = u + at\), where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration (for free fall, this is due to gravity, \(g\)), and \(t\) is the time. When an object is dropped, the initial velocity \(u\) is zero.

For the aforementioned problem, we use Earth’s gravitational acceleration \(a = -9.8 \, \mathrm{m/s^2}\) and the time it took for the ball to reach the ground to calculate the final velocity. The negative sign in \(a\) signifies that gravity acts in the direction opposite to the assumed positive direction (upward). The final velocity just before hitting the ground turns out to be 24.5 m/s downward.

Projectile displacement refers to the change in position of a projectile as it moves along its path; in this case, the vertical distance a ball travels when dropped from a height. This displacement can be calculated using the kinematic equation \(s = ut + \frac{1}{2}at^2\), where \(s\) is the displacement, \(u\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time.

In the context of the building and the ball, the displacement \(s\) corresponds to the height of the building. Given an initial velocity of zero and the acceleration due to gravity, we calculate the building's height by plugging in the time it takes for the ball to reach the ground. Similarly, when estimating the time the ball is visible from a window, we treat the window's lower edge as our point of reference for displacement and resolve for time, keeping in mind the ball's acceleration due to gravity.