Question

The single cable supporting an unoccupied construction elevator breaks when the elevator is at rest at the top of a \(120-\mathrm{m}\) high building. ( \(a\) ) With what speed does the elevator strike the ground? \((b)\) For how long was it falling? \((c)\) What was its speed when it passed the halfway point on the way down? ( \(d\) ) For how long was it falling when it passed the halfway point?

Short answer

The elevator will strike the ground at a speed of approximately \(49.0 m/s\), after falling for roughly \(4.9 s\). It will pass the halfway point (\(60m\)) at a speed of approximately \(34.6 m/s\), and it will take around \(3.5 s\) to get to this point.

Step-by-step solution

Calculation of the final speed at the ground

To determine the speed at which the elevator hits the ground, the kinematic equation, \(v^2 = u^2 + 2gs\) can be used. Where \(v\) is the final velocity, \(u\) is the initial velocity (which is 0 as the elevator starts from rest), \(g\) is the acceleration due to gravity (approximately \(9.8 m/s^2\)), and \(s\) is the distance fallen (which is the height of the building, \(120 m\)). Substituting these values into the equation and solving for \(v\), we get \(v = \sqrt{2gs}\).

Calculation of the time of fall

To calculate the time of fall to the ground, the kinematic equation, \(s = ut + 0.5gt^2\) can be used. Where \(s\) is the distance fallen, \(u\) is the initial velocity, \(g\) is the acceleration due to gravity, and \(t\) is the time of fall. Substituting the known values and solving for \(t\), we can use the square root form as \(t = \sqrt{2s/g}\).

Calculation of the speed at the halfway point

To calculate the speed at which the elevator passes the halfway point, we can use the same method as in step 1, but with the distance \(s\) as half the height of the building (i.e. \(60 m\)). This gives us \(v_{half} = \sqrt{2gs_{half}}\), which we can compute.

Calculation of the time of fall to the halfway point

Lastly, to figure out the time it takes to fall to the halfway point, we can apply the same method as in step 2 but with the distance \(s\) as half the height of the building. This yields \(t_{half} = \sqrt{2s_{half}/g}\) which can be calculated to achieve the specific time.

Key concepts

Free Fall

Free fall refers to the motion of an object under the influence of gravitational force alone. In simpler terms, it's when something falls and only gravity is acting on it, without any air resistance or other forces at play.
When an object is in free fall, such as the construction elevator in the exercise, it starts from rest and accelerates downwards due to gravity. The key point to remember about free fall is the acceleration due to gravity, which allows us to calculate how fast the object is moving and how long it falls.
This is crucial for understanding how the elevator's fall time and impact velocity are derived.

Velocity Calculation

To find out the speed of an object during free fall, we need the initial and final velocities. In this exercise, the elevator begins with an initial velocity (\(u\)) of 0, because it starts from rest at the building's top. By using the formula \(v^2 = u^2 + 2gs\), we can determine the final velocity.

• \(v\) is the final velocity we're solving for
• \(u\) is the initial velocity (0 in this case)
• \(g\) is acceleration due to gravity, which is approximately 9.8 m/s²
• \(s\) is the distance fallen, 120 meters for the elevator

Substituting the values, we use the equation \(v = \sqrt{2gs}\) to find the velocity directly when the elevator hits the ground.
This equation is simple and works well whenever we have an object in free fall.

Kinematic Equations

Kinematic equations are useful for describing the motion of objects. These equations help us calculate variables such as velocity, displacement, and time without needing to know every force.
In our problem, we use two main kinematic equations for free fall:

• To find the final speed: \(v^2 = u^2 + 2gs\)
• To find the time: \(s = ut + 0.5gt^2\)

These equations are derived from the basic principles of motion and assume uniform acceleration, which means gravity is constant and acts in the same way throughout the fall.
The beauty of kinematic equations is their versatility—they allow calculations not just for this specific problem, but for any scenario involving linear motion and constant acceleration.

Acceleration Due to Gravity

Gravity is the force that pulls objects toward the center of the Earth, and its constant acceleration is denoted by \(g\). On the surface of the Earth, this acceleration is approximately 9.8 m/s².
This means that in free fall, every second, an object's velocity increases by 9.8 meters per second. For the elevator, which falls from rest, this constant acceleration makes the velocity increase uniformly until it hits the ground.
Understanding this concept is crucial because it not only contributes to calculating the fall's time duration and final speed but also applies universally to any free-falling object near the Earth's surface.
Whether it's an elevator or a ball, gravity impacts all objects equally, leading to fascinating implications in physics.