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Chapter 2: Problem 70

You are on the roof of the physics building, \(46.0 \mathrm{~m}\) above the ground (Fig. \(\mathbf{P 2 . 7 0}\) ). Your physics professor, who is \(1.80 \mathrm{~m}\) tall, is walking alongside the building at a constant speed of \(1.20 \mathrm{~m} / \mathrm{s}\). If you wish to drop an egg on your professor's head, where should the professor be when you release the egg? Assume that the egg is in free fall.

Short Answer

Expert verified
The professor should be \(3.67 \mathrm{~m}\) away when the egg is released.

Step by step solution

01

Identify relevant quantities

The height the egg is dropped from is \(46.0 \mathrm{~m}\). Because the egg is dropped (as opposed to being thrown), its initial velocity is \(0 \mathrm{~m/s}\). The professor is \(1.80 \mathrm{~m}\) tall and is walking at a constant speed of \(1.20 \mathrm{~m/s}\). The acceleration of the egg is \(9.8 \mathrm{~m/s}^2\) towards the earth due to gravity.
02

Calculate the time for the egg to reach the professor鈥檚 head

We use the equation of motion which is \(h = \frac{1}{2}gt^2\) where \(h\) is the height, \(g\) is the acceleration due to gravity and \(t\) is the time. Solving for \(t\), we get \(t = \sqrt{\frac{2h}{g}}\). Substituting the values we get \(t = \sqrt{\frac{2 \times 46.0}{9.8}} = 3.06 s.\)
03

Calculate the professor鈥檚 distance from the base of the building

Now we need to calculate the distance from the physics building that the professor should be when the egg is released. We can use the formula \(d = vt\) where \(d\) is the distance, \(v\) is the speed and \(t\) is the time. So, on substituting the values we get \(d = 1.20 \times 3.06 = 3.67 m\). Therefore, the professor should be 3.67m away from the base of the building when the egg is released.

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Key Concepts

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

Free Fall
Free fall is a concept in physics that describes the motion of an object subject only to the force of gravity. This implies that there are no other external forces acting on the object, such as air resistance or friction. An object in free fall experiences constant acceleration downward due to gravity. On Earth, this acceleration is approximately \(9.8 \text{ m/s}^2\).
In this problem, the egg is dropped from a height of 46 meters with an initial velocity of zero, which means it begins its descent solely under the influence of gravity. The only force acting on the egg is gravitational force, ensuring it follows the path of free fall. This is crucial for predicting the time it will take for the egg to hit the target, in this case, the professor's head.
Understanding free fall is essential for solving many problems in physics, especially those involving projectile motion. Remember that in a vacuum, without air resistance, all objects near the Earth's surface will fall at the same rate regardless of their mass.
Equations of Motion
The equations of motion are mathematical formulas used to predict the future location or velocity of a moving object. They are vital tools for solving problems in kinematics, which is the study of motion. In this case, we used one such equation to determine the time it takes for the egg to reach the professor's head.
One of the key equations of motion is:
  • \( h = \frac{1}{2}gt^2 \)
where \( h \) represents the height from which the egg is dropped, \( g \) is the acceleration due to gravity, and \( t \) is the time taken to fall. By rearranging this equation to solve for \( t \), we used:
  • \( t = \sqrt{\frac{2h}{g}} \)
This equation helped us compute the time it takes for the egg to descend from a height of 46 meters. These equations are derived from the basic principles of kinematics and allow us to analyze and predict motion in a straightforward manner.
Distance and Velocity Calculation
Distance and velocity calculations are fundamental when analyzing the path of an object in projectile motion. In this problem, after determining the time of fall, we leveraged this to calculate how far the professor should be from the base of the building when the egg is released.
The formula used for this calculation was:
  • \( d = vt \)
Here, \( d \) represents the distance the professor needs to be from the building, \( v \) is the professor's constant walking speed, and \( t \) is the time it took for the egg to fall, which was found to be approximately 3.06 seconds. Plugging in the values, we calculated:
  • \( d = 1.20 \text{ m/s} \times 3.06 \text{ s} = 3.67 \text{ m} \)
These calculations are crucial in determining how motion in the vertical direction (falling egg) affects the position of the moving target (the walking professor) over time. By understanding how to apply these basic calculations, one can solve various problems involving motion.

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Most popular questions from this chapter

During an auto accident, the vehicle's airbags deploy and slow down the passengers more gently than if they had hit the windshield or steering wheel. According to safety standards, airbags produce a maximum acceleration of \(60 g\) that lasts for only \(36 \mathrm{~ms}\) (or less). How far (in meters) does a person travel in coming to a complete stop in \(36 \mathrm{~ms}\) at a constant acceleration of \(60 \mathrm{~g}\) ?

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

A small block has constant acceleration as it slides down a frictionless incline. The block is released from rest at the top of the incline, and its speed after it has traveled \(6.80 \mathrm{~m}\) to the bottom of the incline is \(3.80 \mathrm{~m} / \mathrm{s}\). What is the speed of the block when it is \(3.40 \mathrm{~m}\) from the top of the incline?

A \(7500 \mathrm{~kg}\) rocket blasts off vertically from the launch pad with a constant upward acceleration of \(2.25 \mathrm{~m} / \mathrm{s}^{2}\) and feels no appreciable air resistance. When it has reached a height of \(525 \mathrm{~m}\), 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.

A student is running at her top speed of \(5.0 \mathrm{~m} / \mathrm{s}\) to catch a bus, which is stopped at the bus stop. When the student is still \(40.0 \mathrm{~m}\) from the bus, it starts to pull away, moving with a constant acceleration of \(0.170 \mathrm{~m} / \mathrm{s}^{2}\). (a) For how much time and what distance does the student have to run at \(5.0 \mathrm{~m} / \mathrm{s}\) before she overtakes the bus? (b) When she reaches the bus, how fast is the bus traveling? (c) Sketch an \(x-t\) graph for both the student and the bus. Take \(x=0\) at the initial position of the student. (d) The equations you used in part (a) to find the time have a second solution, corresponding to a later time for which the student and bus are again at the same place if they continue their specified motions. Explain the significance of this second solution. How fast is the bus traveling at this point? (e) If the student's top speed is \(3.5 \mathrm{~m} / \mathrm{s},\) will she catch the bus? (f) What is the minimum speed the student must have to just catch up with the bus? For what time and what distance does she have to run in that case?
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