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Chapter 4: Problem 17

A steel ball is dropped from the roof of a building. A man standing in front of a \(1-m\) high window in the building notes that the ball takes \(0.1\) s to the fall from the top to the bottom of the window. The ball continues to fall and strikes the ground. On striking the ground, the ball gets rebounded with the same speed with which it hits the ground. If the ball reappears at the bottom of the window \(2 \mathrm{~s}\) after passing the bottom of the window on the way down, find the height of the building.

Short Answer

Expert verified
The height of the building is 15.9 meters.

Step by step solution

01

Understanding the Problem

First, analyze what is happening. The steel ball falls past a window and later reappears at the same point after bouncing off the ground. We need to calculate the height of the building from which it was dropped.
02

Determine the Speed Past the Window

The ball falls past a window that is 1 meter high in 0.1 seconds. We can calculate the speed of the ball as it passes the window using the formula for constant velocity, since window height is relatively small compared to the whole fall:\[ v = \frac{d}{t} = \frac{1 \, \text{m}}{0.1 \, \text{s}} = 10 \, \text{m/s} \]
03

Calculate Speed Upon Hitting the Ground

The ball reappears at the bottom of the window 2 seconds after passing it on the way down. This total time includes falling after passing the window, touching the ground, and rebounding back to the window. Thus, when it returns, it maintains the same speed just after rebounding: \[ v = 10 \, \text{m/s} \]
04

Calculate Time of Fall from the Window to the Ground

Since the total descent and ascent time is 2 seconds, the actual time to descend and hit the ground must be half of this (as rebound is equal in time and speed after hitting): \[ t = \frac{2}{2} = 1 \, \text{s} \]
05

Use Kinematics to Find Height of the Rest of the Building

Using the kinematic equation for initial speed and time, we find the distance from the window to the ground:\[ h = v \cdot t + \frac{1}{2} g t^2 \] Plug the values (\( v = 10 \, \text{m/s} \), \( t = 1 \, \text{s} \), \( g = 9.8 \, \text{m/s}^2 \)):\[ h = 10 \cdot 1 + \frac{1}{2} \cdot 9.8 \cdot 1^2 \] Calculate:\[ h = 10 + 4.9 = 14.9 \, \text{m} \]
06

Calculate Total Building Height

Add the height of the window to the calculated height from window base to ground:\[ H = 14.9 \, \text{m} + 1 \, \text{m} = 15.9 \, \text{m} \]
07

Conclusion

The total height of the building is thus \( 15.9 \, \text{m} \).

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Key Concepts

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

Free Fall
When an object is in free fall, it essentially means the object is only under the influence of gravity. This is what happens to a steel ball dropped from a building roof. As soon as the ball is released, it accelerates downwards because of Earth's gravitational pull.

The fall is 'free' in the sense that the only force acting on the ball is gravity 鈥� other forces like air resistance are either negligible or not considered in basic kinematic equations.
  • For an object in free fall near the Earth's surface: the acceleration due to gravity is typically approximately \(9.8 \ m/s^2\).
  • Objects in free fall don't start from a zero velocity; they may have an initial velocity if they begin moving.
Understanding free fall is vital when working through problems involving objects dropping from heights and calculating their velocity or the height they fall from.
Velocity Calculation
Velocity gives us an idea of the speed and direction of a moving object. When calculating the velocity as a ball passes by a window during its free fall, we use a simple formula for constant flux.

The velocity calculation involves dividing the distance by the time taken. Here, the distance is the height of the window (1 m), and the time is how long the ball takes to pass this window, which is 0.1 seconds.
  • The formula is: \( v = \frac{d}{t} \), where \(v\) is velocity, \(d\) is distance, and \(t\) is time.
  • In the given problem, \( v = \frac{1\ m}{0.1\ s} = 10\ m/s\), indicating the ball is traveling at 10 m/s when it is at the level of the window.
Calculating velocity this way provides valuable insights into the object鈥檚 motion at specific points during its descent.
Kinematic Equations
Kinematic equations are used to describe the motion of objects without considering the causes of motion. In problems involving free fall like the current one, they give critical insights on distances and velocities.

There are several kinematic equations, but a common one for figuring out the height an object falls under constant acceleration (like gravity) is:
  • \( h = v \cdot t + \frac{1}{2} g t^2 \)
Here, \( h \) is the height, \( v \) is the initial velocity (velocity at the moment of passing the window for the drop), \( g \) is gravitational acceleration, and \( t \) is time.

In the problem, passing the window the ball has an initial velocity of 10 m/s, and it takes 1 second to reach the ground, contributing to the final calculation of the building鈥檚 height.
  • The combination of kinematic equations and measured time, allows us to piece together information leading to a conclusion about the distance or in this case, building height.
Kinematic equations simplify the calculation of motion aspects, which can be applied broadly across different types of motion.

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Most popular questions from this chapter

Check up the only correct statements in the following: (1) A body having a constant velocity still can have varying speed. (2) A body having a constant speed can have varying velocity. (3) A body having constant speed can have an acceleration. (4) If velocity and acceleration are in the same direction, then distance is equal to displacement.

A police party is chasing a dacoit in a jeep which is moving at a constant speed \(v\). The dacoit is on a motorcycle. When he is at a distance \(x\) from the jeep, he accelerates from rest at a constant rate. Which of the following relations is true if the police is able to catch the dacoit? (1) \(v^{2} \leq \infty\) (2) \(v^{2} \leq 2 \alpha x\) (3) \(v^{2} \geq 2 \omega x\) (4) \(v^{2} \geq \alpha x\)

A body is projected upwards with a velocity \(u\). It passes through a certain point above the ground after \(t_{1}\). The time afler which the body passes through the same point during the retum journey is (1) \(\left(\frac{u}{g}-t_{1}^{2}\right)\) (2) \(2\left(\frac{u}{g}-t_{1}\right)\) (3) \(3\left(\frac{u^{2}}{g}-t_{1}\right)\) (4) \(3\left(\frac{u^{2}}{g^{2}}-t_{1}\right)\)

\(B_{1}, B_{2}\), and \(B_{3}\) are three balloons ascending with velocities \(v, 2 v\), and \(3 v\), respectively. If a bomb is dropped from each when they are at the same height, then (1) Bomb from \(B_{1}\) reaches ground first (2) Bomb from \(B_{2}\) reaches ground first (3) Bomb from \(B_{3}\) reaches ground first (4) They reach the ground simultaneously

A particle is moving along the \(x\)-axis whose instantaneous speed is given by \(v^{2}=108-9 x^{2}\). The acceleration of the particle is (1) \(-9 x \mathrm{~ms}^{-2}\) (2) \(-18 x \mathrm{~ms}^{-2}\) (3) \(\frac{-9 x}{2} \mathrm{~ms}^{-2}\) (4) None of these
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