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Chapter 4: Problem 9

An object is dropped from the top of a building and is in view for time \(\tau\) while passing a window of height \(h\) some distance lower down. How high is the top of the building above the top of the window?

Short Answer

Expert verified
The height of the top of the building above the top of the window is given by \(H - h\), where \(H = 0.5gT^2\) and \(h = 0.5g\tau^2\). Here, \(T\) is the total time of fall and can be given by \(\tau\), the time it takes the object to pass the window.

Step by step solution

01

Identify relevant information

We have two points of interest here. The point from which the object is dropped, and the top point of the window. Let's denote the total height of the building from the point of the object dropped as \(H\), and the height of the window from the ground as \(h\). We're given the time it takes for the object to pass the window, \(\tau\), but not the total time the object falls. Let's denote it as \(T\).
02

Apply equations of motion to the window

The object falls through the window in time \(\tau\). This means we can apply the equation of motion to find the height of the window, \(h = 0.5g\tau^2\). Here, \(g\) is the acceleration due to gravity.
03

Apply equations of motion to the whole building

We know the total height of the building is \(H\), but we don't know the total time of fall, \(T\). But we do know that the final velocity when the object passes the top of the window is \(v = g\tau\). Plug this value into the second equation to find \(T\): \(v = gT\), or \(T = v/g\).
04

Find the height H by using the total time T

We have the total time of fall \(T\), so we can plug it into the first equation to find the total height \(H\): \(H = 0.5gT^2\). Replace \(T\) by the value from Step 3 (i.e. \(\tau\)) to find \(H\).
05

Deduce the height between the drop point and the top of the window

We now know the total height of the building (\(H\)) and the height of the window (\(h\)). The difference between these two values is the height from the top of the building to the top of the window: \(H - h\).

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Key Concepts

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

Free Fall Physics
When we think about an object falling freely under the influence of gravity, we are diving into the realm of free fall physics. This concept assumes that the only force acting on the object is gravity, ignoring air resistance and other forces. It's a simplified model that comes in handy when solving physics problems.

In free fall, all objects accelerate towards the Earth at the same rate, regardless of their mass. This uniform acceleration is due to the pull of Earth's gravity. If you were to drop a feather and a hammer in a vacuum (where there's no air resistance), they would hit the ground simultaneously. This astonishing phenomenon was famously demonstrated by Apollo 15 astronaut David Scott on the Moon, where there is no atmosphere to interfere with free fall.

The time it takes an object to fall and the distance it covers are directly related to this constant acceleration, leading us to the next core concept - kinematic equations, which help us analyze motion in scenarios like the one in our exercise.
Kinematic Equations
The kinematic equations are the tools we use to describe an object's motion, including its velocity, acceleration, and position as functions of time. These equations are especially useful for problems involving constant acceleration, like free fall.

For example, the equation \( s = ut + \frac{1}{2}at^2 \) relates the distance \( s \) moved in a certain time \( t \) with the initial velocity \( u \) and the constant acceleration \( a \) of the object. When we're dealing with free fall from rest (where initial velocity is zero), this simplifies to \( s = \frac{1}{2}gt^2 \), where \( g \) is the acceleration due to gravity. This is the equation used in Step 2 of our example to calculate the height of the window the object falls past.

Understanding these equations allows us to create a mathematical model of an object's fall and predict its position at any given time, which is essential when there is no direct measurement of the fall duration, as illustrated in the textbook exercise.
Acceleration Due to Gravity
The constant \( g \) we often see in kinematic equations stands for the acceleration due to gravity at the Earth's surface, and it has an average value of approximately \( 9.81 \text{ m/s}^2 \). This value can slightly vary depending on altitude and geographical location, but for most practical purposes, we consider it constant.

Gravity is what gives weight to physical objects and causes them to fall toward the ground when dropped. Any object experiencing free fall is accelerating downwards at this rate, hence the term 'acceleration due to gravity'. This acceleration is the sole reason why an object in free fall does not move at a constant velocity, but instead, its speed increases every second by approximately \( 9.81 \text{ m/s} \).

When solving problems related to free fall, it's essential to understand that this acceleration is the driving factor behind an object's changing velocity and position. So when we encounter an exercise like the one provided, recognising that we can find the height from which an object is dropped by calculating the object's free fall time and applying kinematic equations, involves the underlying concept of acceleration due to gravity.

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Most popular questions from this chapter

A particle \(P\) of mass \(m\) can slide along a smooth rigid straight wire. The wire has one of its points fixed at the origin \(O\), and is made to rotate in the \((x, y)\)-plane with angular speed \(\Omega\). By using the vector equation of motion of \(P\) in polar co-ordinates, show that \(r\), the distance of \(P\) from \(O\), satisfies the equation $$\ddot{r}-\Omega^{2} r=0$$ and find a second equation involving \(N\), where \(N \widehat{\theta}\) is the force the wire exerts on \(P\). [Ignore gravity in this question.] Initially, \(P\) is at rest (relative to the wire) at a distance \(a\) from \(O\). Find \(r\) as a function of \(t\) in the subsequent motion, and deduce the corresponding formula for \(N\).

A mortar gun, with a maximum range of \(40 \mathrm{~m}\) on level ground, is placed on the edge of a vertical cliff of height \(20 \mathrm{~m}\) overlooking a horizontal plain. Show that the horizontal range \(R\) of the mortar gun is given by $$R=40\left\\{\sin \alpha+\left(1+\sin ^{2} \alpha\right)^{\frac{1}{2}}\right\\} \cos \alpha$$ where \(\alpha\) is the angle of elevation of the mortar above the horizontal. [Take \(g=10 \mathrm{~m} \mathrm{~s}^{-2}\).] Evaluate \(R\) (to the nearest metre) when \(\alpha=45^{\circ}\) and \(35^{\circ}\) and confirm that \(\alpha=45^{\circ}\) does not yield the maximum range. [Do not try to find the optimum projection angle this way. See Problem \(4.22\) below. \(]\)

A particle \(P\) of mass \(m\) moves under the gravitational attraction of a mass \(M\) fixed at the origin \(O\). Initially \(P\) is at a distance \(a\) from \(O\) when it is projected with the critical escape speed \((2 M G / a)^{1 / 2}\) directly away from \(O\). Find the distance of \(P\) from \(O\) at time \(t\), and confirm that \(P\) escapes to infinity.

A microscopic spherical oil droplet, of density \(\rho\) and unknown radius, carries an unknown electric charge. The droplet is observed to have terminal speed \(v_{1}\) when falling vertically in air of viscosity \(\mu\). When a uniform electric field \(E_{0}\) is applied in the vertically upwards direction, the same droplet was observed to move upwards with terminal speed \(v_{2}\). Find the charge on the droplet. [Use the low Reynolds number approximation for the drag.]

A particle of mass \(m\) can move on a rough horizontal table and is attached to a fixed point on the table by a light inextensible string of length \(b\). The resistance force exerted on the particle is \(-m K v\), where \(v\) is the velocity of the particle. Initially the string is taut and the particle is projected horizontally, at right angles to the string, with speed \(u\). Find the angle turned through by the string before the particle comes to rest. Find also the tension in the string at time \(t\).
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