Question

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?

Short answer

b) What is the height of the building? c) At what time is the ball first seen in the window that is 2.5m above the ground and has a height of 1.2m?

Step-by-step solution

Find the time taken for the ball to reach the ground

The ball takes the same amount of time to fall from the roof to the ground and then back to the original height. Hence, the time taken for the ball to reach the ground is half of the total time. So, we find the time taken for the ball to reach the ground as: $$t_{ground}=\frac{5.0}{2}=2.5\mathrm{s}$$

Calculate the speed of the ball just before it hits the ground

Using the kinematic equation $v=u+at$, where $v$ is the final velocity, $u$ is the initial velocity (0 m/s), $a$ is the acceleration due to gravity ($-9.8\mathrm{m}/{\mathrm{s}}^2$), and $t$ is the time taken for the ball to reach the ground, we have: $$v=0-(9.8)(2.5)=-24.5\mathrm{m}/\mathrm{s}$$ The negative sign indicates that the velocity is downward. So, the speed of the ball just before it hits the ground is 24.5 m/s.

Calculate the height of the building

Using the kinematic equation $s=ut+\frac{1}{2}a{t}^2$, where $s$ is the displacement (the height of the building), $u$ is the initial velocity (0 m/s), $a$ is the acceleration due to gravity ($-9.8\mathrm{m}/{\mathrm{s}}^2$), and $t$ is the time taken for the ball to reach the ground, we have: $$s=0(2.5)+\frac{1}{2}(-9.8)(2.5{)}^2=-30.625\mathrm{m}$$ The negative sign indicates that the displacement is downward. So, the height of the building is 30.625 m.

Calculate the time when the ball is first seen in the window

Let's denote the time when the ball is first seen in the window as $t_{window}$. We know that the window is 2.5m above the ground, and the height of the window is 1.2 m. So, the ball is first seen in the window at a height of $2.5+1.2=3.7\mathrm{m}$ above the ground. Using the equation $s=ut+\frac{1}{2}a{t}^2$, and remembering that $u$ is 0 m/s and $a$ is $-9.8\mathrm{m}/{\mathrm{s}}^2$, we have: $$3.7=0(t_{window})+\frac{1}{2}(-9.8)(t_{window}{)}^2$$ Solving for $t_{window}$, we get: $$t_{window}=\sqrt{\frac{2(3.7)}{9.8}}\approx \sqrt{0.753}\approx 0.87\mathrm{s}$$ So, the ball is first seen in the window approximately 0.87 seconds after it was dropped.

Key concepts

Free Fall Motion

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 $9.8\mathrm{m}/{\mathrm{s}}^2$.

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.

Final Velocity Calculation

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 $a=-9.8\mathrm{m}/{\mathrm{s}}^2$ 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

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 $s=ut+\frac{1}{2}a{t}^2$, 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.