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Chapter 3: Problem 24

A criminal is escaping across a rooftop and runs off the roof horizontally at a speed of \(5.3 \mathrm{m} / \mathrm{s}\), hoping to land on the roof of an adjacent building. Air resistance is negligible. The horizontal distance between the two buildings is \(D,\) and the roof of the adjacent building is \(2.0 \mathrm{m}\) below the jumping-off point. Find the maximum value for \(D\).

Short Answer

Expert verified
The maximum horizontal distance he can travel is approximately 3.39 meters.

Step by step solution

01

Understanding the Problem

The problem involves a projectile motion where the criminal runs off a rooftop horizontally. The initial vertical velocity is 0, and the building he lands on is 2 meters below the starting point. We need to determine the maximum horizontal distance (D) he can travel given these conditions.
02

Calculate Time of Fall

We need to calculate the time it takes for the criminal to fall 2 meters vertically. Using the formula for vertical motion under gravity: \[ y = v_{iy} \cdot t + \frac{1}{2} g t^2 \]where \( y = 2 \,\mathrm{m} \), \( v_{iy} = 0 \ \,\mathrm{m/s} \) (initial vertical velocity), and \( g = 9.8 \,\mathrm{m/s^2} \) (acceleration due to gravity), we solve for \( t \):\[ 2 = \frac{1}{2} \times 9.8 \times t^2 \]\[ 2 = 4.9t^2 \]\[ t^2 = \frac{2}{4.9} \approx 0.408 \]\[ t \approx 0.639 \,\mathrm{s} \].
03

Calculate Maximum Horizontal Distance

Since air resistance is negligible, the horizontal motion is constant. The horizontal velocity is given as \( 5.3 \,\mathrm{m/s} \). Using the formula for horizontal distance \( D = v_{x} \cdot t \), where \( v_{x} = 5.3 \,\mathrm{m/s} \) and \( t = 0.639 \,\mathrm{s} \):\[ D = 5.3 \times 0.639 \approx 3.39 \,\mathrm{m} \].

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Key Concepts

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

Horizontal Distance Calculation
The horizontal distance a projectile travels is determined by its initial horizontal velocity and the time it spends in the air. In this scenario, the criminal runs off the roof at a speed of 5.3 m/s horizontally. This means there is no horizontal acceleration since air resistance is negligible.
To find the maximum horizontal distance (D), we use the formula:
  • \( D = v_{x} \times t \)
  • where \( v_{x} = 5.3 \, \text{m/s} \) and \( t \) is the time of fall.
This means that the horizontal distance is directly proportional to the time the criminal spends in the air. The longer the fall time, the further the horizontal distance covered. Hence, accurately calculating the time of fall is crucial in determining the trajectory reach.
Vertical Motion Under Gravity
Vertical motion under gravity is the key factor in determining how long the criminal will be airborne. The formula that governs this is:
  • \( y = v_{iy} \cdot t + \frac{1}{2} g t^2 \)
  • where \( y \) is the vertical displacement, \( v_{iy} \) is the initial vertical velocity, \( g \) is the gravitational acceleration (9.8 m/s虏), and \( t \) is the time.
In this problem, because the jump is horizontal, the initial vertical velocity \( v_{iy} \) is 0 m/s. Therefore, the motion only involves the gravitational pull:
  • \( y = \frac{1}{2} g t^2 \)
With a downward displacement of 2 m, the formula helps us compute the time of fall by rearranging it to solve for \( t \). This physics principle allows us to factor in only gravitational effects, simplifying the calculation for practical use and predictions.
Time of Fall Calculation
Calculating the time it takes for the projectile to fall is an essential part of solving projectile motion problems. By determining this, we can find out how long the criminal remains in the air, which, in turn, dictates the horizontal distance traveled.
To find this time:
  • We use \( 2 = \frac{1}{2} \times 9.8 \times t^2 \).
  • After solving \( 2 = 4.9t^2 \), we find \( t^2 = \frac{2}{4.9} \).
  • This gives \( t^2 \approx 0.408 \), leading to \( t \approx 0.639 \, \text{s} \).
This time calculation is crucial since it directly affects the horizontal distance calculation. Without accurately measuring the fall time, the prediction of the jump's range would be incorrect. Understanding these steps ensures proper assessment of similar real-world scenarios.

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Most popular questions from this chapter

In a marathon race Chad is out in front, running due north at a speed of \(4.00 \mathrm{m} / \mathrm{s} .\) John is \(95 \mathrm{m}\) behind him, running due north at a speed of \(4.50 \mathrm{m} / \mathrm{s} .\) How long does it take for John to pass Chad?

A volleyball is spiked so that it has an initial velocity of \(15 \mathrm{m} / \mathrm{s}\) directed downward at an angle of \(55^{\circ}\) below the horizontal. What is the horizontal component of the ball's velocity when the opposing player fields the ball?

On a pleasure cruise a boat is traveling relative to the water at a speed of \(5.0 \mathrm{m} / \mathrm{s}\) due south. Relative to the boat, a passenger walks toward the back of the boat at a speed of \(1.5 \mathrm{m} / \mathrm{s} .\) (a) What are the magnitude and direction of the passenger's velocity relative to the water? (b) How long does it take for the passenger to walk a distance of \(27 \mathrm{m}\) on the boat? (c) How long does it take for the passenger to cover a distance of \(27 \mathrm{m}\) on the water?

At some airports there are speed ramps to help passengers get from one place to another. A speed ramp is a moving conveyor belt on which you can either stand or walk. Suppose a speed ramp has a length of \(105 \mathrm{m}\) and is moving at a speed of \(2.0 \mathrm{m} / \mathrm{s}\) relative to the ground. In addition, suppose you can cover this distance in 75 s when walking on the ground. If you walk at the same rate with respect to the speed ramp that you walk on the ground, how long does it take for you to travel the \(105 \mathrm{m}\) using the speed ramp?

A jetliner can fly 6.00 hours on a full load of fuel. Without any wind it flies at a speed of \(2.40 \times 10^{2} \mathrm{m} / \mathrm{s} .\) The plane is to make a round-trip by heading due west for a certain distance, turning around, and then heading due east for the return trip. During the entire flight, however, the plane encounters a \(57.8-\mathrm{m} / \mathrm{s}\) wind from the jet stream, which blows from west to east. What is the maximum distance that the plane can travel due west and just be able to return home?
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