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Chapter 2: Problem 74

A flowerpot falls off a windowsill and passes the window of the story below. Ignore air resistance. It takes the pot 0.380 s to pass from the top to the bottom of this window, which is 1.90 m high. How far is the top of the window below the windowsill from which the flowerpot fell?

Short Answer

Expert verified
The top of the window is 0.5025 m below the windowsill.

Step by step solution

01

Understand the Problem

We need to find how far the top of the window is below the windowsill from which the flowerpot fell. We are given the time it takes for the flowerpot to pass a window with a height of 1.90 m. Use the equations of uniformly accelerated motion to solve this.
02

Identify Relevant Equations

We will use the kinematic equation for uniform acceleration: \[ s = ut + \frac{1}{2} a t^2 \]where \( s \) is the distance traveled, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time. The acceleration \( a \) is the acceleration due to gravity \( 9.8 \text{ m/s}^2 \), and the window height \( 1.90 \text{ m} \).
03

Calculate Initial Velocity at the Top of the Window

The flowerpot has an initial velocity \( u \) when its top reaches the top of the window. To find this, we rearrange and use the window passage: \[ s = ut + \frac{1}{2} a t^2 \]Plugging in \( s = 1.90 \text{ m} \), \( t = 0.380 \text{ s} \), and \( a = 9.8 \text{ m/s}^2 \), we solve for \( u \):\[ 1.90 = u(0.380) + \frac{1}{2} \times 9.8 \times (0.380)^2 \] \[ 1.90 = 0.38u + 0.70764 \]\[ 0.38u = 1.9 - 0.70764 \]\[ 0.38u = 1.19236 \]\[ u \approx 3.1383 \text{ m/s} \]
04

Calculate the Height from the Windowsill to Top of the Window

Now, use \( u = 3.1383 \text{ m/s} \) to find the distance from the windowsill. Originally the flowerpot started from rest, and using the same kinematic equation:\[ u^2 = 2as \]\[ (3.1383)^2 = 2 \times 9.8 \times s \]\[ s = \frac{3.1383^2}{2 \times 9.8} \]\[ s = \frac{9.84796}{19.6} \]\[ s \approx 0.5025 \text{ m} \]
05

Conclusion

The distance from the windowsill to the top of the window is approximately 0.5025 m.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Uniform Acceleration
Uniform acceleration occurs when an object's velocity changes at a consistent rate. In the context of our exercise, the flowerpot falling from the windowsill undergoes a uniform acceleration due to gravity. This means that the pot speeds up steadily as it descends toward the ground.

The constant acceleration due to gravity on Earth is approximately 9.8 m/s虏. This value is critical for calculating how far an object falls, influencing both its speed and position over time. By knowing this constant, we can predict the future position of any object in free fall, assuming there is no air resistance slowing it down.

In problems involving uniform acceleration like this one, we rely heavily on equations that relate time, distance, initial velocity, and acceleration to understand the movement of objects. Understanding uniform acceleration provides insight into how and why objects in free fall behave the way they do.
Kinematic Equation
Kinematic equations are mathematical formulas used in physics to describe the motion of objects. They relate various factors like displacement (\( s \)), initial velocity (\( u \)), time (\( t \)), and acceleration (\( a \)).

In our flowerpot problem, we use the kinematic equation \( s = ut + \frac{1}{2} a t^2 \) to determine different parameters of motion. This specific equation is useful when dealing with uniformly accelerated motion. It helps find the distance traveled or the initial velocity, as done in solving the exercise.

This equation assumes no other forces act on the object except for uniform acceleration 鈥� in this case, gravity. It is crucial to carefully substitute the known values into the equation to solve for the unknown, like the height from which the flowerpot fell. Becoming proficient at using kinematic equations allows you to tackle any motion-related problem with confidence.
Free Fall
Free fall describes the motion of an object when gravity is the only force acting upon it. In our exercise, the flowerpot is in free fall once it leaves the windowsill. This means it accelerates downward at that steady rate of 9.8 m/s虏, which is the gravitational acceleration.

This condition is ideal for applying kinematic equations. Ignoring air resistance allows us to simplify the problem, focusing only on the gravitational pull. When we consider the time it took the flowerpot to travel past the window, it helps us understand how free fall works in practical scenarios.

Being able to calculate such movements is fundamental in physics to predict how long it will take objects to hit the ground, how fast they will be traveling upon impact, and the distance they will cover during their fall. Understanding free fall is key to mastering basic kinematics and grasping the underlying principles of motion.

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Most popular questions from this chapter

\(\textbf{Prevention of Hip Fractures.}\) Falls resulting in hip fractures are a major cause of injury and even death to the elderly. Typically, the hip鈥檚 speed at impact is about 2.0 m/s. If this can be reduced to 1.3 m/s or less, the hip will usually not fracture. One way to do this is by wearing elastic hip pads. (a) If a typical pad is 5.0 cm thick and compresses by 2.0 cm during the impact of a fall, what constant acceleration (in m/s\(^{2}\) and in \(\text{g}\)鈥檚) does the hip undergo to reduce its speed from 2.0 m/s to 1.3 m/s? (b) The acceleration you found in part (a) may seem rather large, but to assess its effects on the hip, calculate how long it lasts.

You are standing at rest at a bus stop. A bus moving at a constant speed of \(5.00 \mathrm{~m} / \mathrm{~s}\) passes you. When the rear of the bus is \(12.0 \mathrm{~m}\) past you, you realize that it is your bus, so you start to run toward it with a constant acceleration of \(0.960 \mathrm{~m} / \mathrm{~s}^{2}\). How far would you have to run before you catch up with the rear of the bus, and how fast must you be running then? Would an average college student be physically able to accomplish this?

An object is moving along the \(x\)-axis. At \(t =\) 0 it has velocity \(v_{0x}\) = 20.0 m/s. Starting at time \(t =\) 0 it has acceleration \(a_x = -Ct\), where \(C\) has units of m/s\(^3\). (a) What is the value of \(C\) if the object stops in 8.00 s after \(t =\) 0? (b) For the value of \(C\) calculated in part (a), how far does the object travel during the 8.00 s?

A 15-kg rock is dropped from rest on the earth and reaches the ground in 1.75 s. When it is dropped from the same height on Saturn's satellite Enceladus, the rock reaches the ground in 18.6 s. What is the acceleration due to gravity on Enceladus?

A 7500-kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.25 m/s\(^2\) and feels no appreciable air resistance. When it has reached a height of 525 m, its engines suddenly fail; the only force acting on it is now gravity. (a) What is the maximum height this rocket will reach above the launch pad? (b) How much time will elapse after engine failure before the rocket comes crashing down to the launch pad, and how fast will it be moving just before it crashes? (c) Sketch \(a_y-t, v_y-t\), and \(y-t\) graphs of the rocket's motion from the instant of blast-off to the instant just before it strikes the launch pad.
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