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Chapter 2: Problem 63

A drowsy cat spots a flowerpot that sails first up and then down past an open window. The pot is in view for a total of \(0.50 \mathrm{~s}\), and the top-to- bottom height of the window is \(2.00 \mathrm{~m}\). How high above the window top does the flowerpot go?

Short Answer

Expert verified
The flowerpot rises approximately 0.53 m above the window top.

Step by step solution

01

Define the Problem

We are given the time the flowerpot is visible and the height of the window. We need to find the maximum height the flowerpot reaches above the window top.
02

Analyze the Motion

The flowerpot's motion can be split into two parts: ascending to the highest point and descending back to the bottom of the window. The total visible time is 0.50 s for both upward and downward motion.
03

Calculate the Time to Reach Max Height

Since the total visible time is 0.50 s for both ascending and descending, the time taken to reach the maximum height just above the window will be half of the time it is in view. So, calculate the time to reach from the bottom to the top or vice versa as: \ \[ t_{one \, way} = \frac{0.50}{2} = 0.25 \, s \]
04

Use Kinematics to Find Initial Velocity

Use the kinematic equation for the motion from the bottom to the top of the window for half the visible time: \ \[ s = v_0 \times t + \frac{1}{2} a t^2 \] \ where \( s = 2.00 \, m \), \( t = 0.25 \, s \), and acceleration \( a = -9.81 \, m/s^2 \).\ The equation becomes \ \[ 2.00 = v_0 \times 0.25 - \frac{1}{2} \times 9.81 \times (0.25)^2 \] \ Solving for \( v_0 \), the initial velocity needed to reach the top of the window is found.
05

Calculate Maximum Height Above Window

Once the initial velocity \( v_0 \) is found, use \( v_f^2 = v_0^2 + 2a h_{above} \) where \( v_f = 0 \) at the maximum height to find the distance above the window top: \ \[ h_{above} = \frac{v_0^2}{2 \times 9.81} \].\
06

Find the Final Result

Compute the value of \( h_{above} \) using the initial velocity from Step 4 to find how high above the window top the flowerpot goes.

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Key Concepts

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

Kinematics
Kinematics is essential for understanding how objects move, without considering the forces that cause the motion. It involves breaking down the motion into simpler parts to analyze each phase. For example, when analyzing a flowerpot's motion past a window, the problem is divided into ascent and descent.
  • On its way up, the pot slows down due to gravity until it reaches its highest point (maximum height).
  • On its way down, the pot speeds up as it returns past the window.
Kinematic equations are invaluable tools for solving these problems because they link initial velocity, final velocity, acceleration, displacement, and time. These equations allow us to determine unknowns like the maximum height the pot reaches above the window simply by plugging in known values.
Free Fall
Free fall is a special type of motion in physics where an object is only acted upon by the force of gravity. This means there are no other forces, like air resistance, affecting it. A flowerpot in free fall will experience a constant acceleration downward, which is why it decelerates on the way up and accelerates on the way down.During its entire motion:
  • The acceleration remains constant at approximately \( g = 9.81 \, \text{m/s}^2 \)
  • Even when the pot is momentarily stationary at its peak, it's still accelerating due to gravity (ready to fall back down).
Understanding free fall helps predict the complete motion path and timing, ensuring we know how long the pot is visible through a window and how high it ascends.
Acceleration due to Gravity
Acceleration due to gravity is a crucial constant in projectile motion, symbolized by \( g \), which on Earth is approximately \( 9.81 \, \text{m/s}^2 \). This constant value means that every second, an object's velocity changes by \( 9.81 \, \text{m/s} \) towards the Earth's surface; this influence is always directed downwards.To find how high above the window the flowerpot reaches, this constant is key. By knowing the initial velocity (found through kinematics), we can use formulas like \( v_f^2 = v_0^2 + 2a h_{above} \), where \( v_f = 0 \) at the maximum height, to calculate the extra height above the window the pot achieves. Understanding how the acceleration due to gravity affects both the ascent and descent offers insights into practical applications, such as timing in air traffic or the development of sports equipment.

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Most popular questions from this chapter

A rock is shot vertically upward from the edge of the top of a tall building. The rock reaches its maximum height above the top of the building \(1.60 \mathrm{~s}\) after being shot. Then, after barely missing the edge of the building as it falls downward, the rock strikes the ground \(6.00 \mathrm{~s}\) after it is launched. In SI units: (a) with what upward velocity is the rock shot, (b) what maximum height above the top of the building is reached by the rock, and (c) how tall is the building?

A ball is thrown vertically downward from the top of a 36.6m-tall building. The ball passes the top of a window that is \(12.2 \mathrm{~m}\) above the ground \(2.00 \mathrm{~s}\) after being thrown. What is the speed of the ball as it passes the top of the window?

The position of a particle moving along the \(x\) axis is given in centimeters by \(x=9.75+1.50 t^{3}\), where \(t\) is in seconds. Calculate (a) the average velocity during the time interval \(t=2.00 \mathrm{~s}\) to \(t=3.00 \mathrm{~s}\); (b) the instantaneous velocity at \(t=2.00 \mathrm{~s} ;(\mathrm{c})\) the instantaneous velocity at \(t=3.00 \mathrm{~s} ;\) (d) the instantaneous velocity at \(t=2.50 \mathrm{~s}\); and (e) the instantaneous velocity when the particle is midway between its positions at \(t=2.00 \mathrm{~s}\) and \(t=3.00 \mathrm{~s}\). (f) Graph \(x\) versus \(t\) and indicate your answers graphically.

On a dry road, a car with good tires may be able to brake with a constant deceleration of \(4.92 \mathrm{~m} / \mathrm{s}^{2} .\) (a) How long does such a car, initially traveling at \(24.6 \mathrm{~m} / \mathrm{s}\), take to stop? (b) How far does it travel in this time? (c) Graph \(x\) versus \(t\) and \(v\) versus \(t\) for the deceleration.

A jumbo jet must reach a speed of \(360 \mathrm{~km} / \mathrm{h}\) on the runway for takeoff. What is the lowest constant acceleration needed for takeoff from a \(1.80 \mathrm{~km}\) runway?
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