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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 \( D \) is approximately \( 3.38 \) meters.

Step by step solution

01

Analyze the situation and components of motion

The criminal is jumping horizontally with no initial vertical velocity. There are two components of motion: horizontal and vertical. The horizontal motion is uniform, so the horizontal speed remains constant at \(5.3\,\text{m/s}\). The vertical motion is accelerated under the influence of gravity \( (g = 9.8\,\text{m/s}^2) \).
02

Determine the time to fall 2.0 m

Using the equation for the vertical motion due to gravity: \[ y = v_{iy}t + \frac{1}{2}gt^2 \] where \(y = 2.0\,\text{m}\), \(v_{iy} = 0\,\text{m/s}\), and \(g = 9.8\,\text{m/s}^2\). Rearranging for \( t \), we solve:\[ 2.0 = \frac{1}{2} \times 9.8 \times t^2 \]\[ t^2 = \frac{4.0}{9.8} \approx 0.408 \]\[ t = \sqrt{0.408} \approx 0.639\,\text{s} \]
03

Determine maximum horizontal distance D

Using the uniform horizontal motion equation, where horizontal speed \(v_{x} = 5.3\,\text{m/s}\) and time \(t = 0.639\,\text{s}\):\[ D = v_{x} \times t \]\[ D = 5.3 \times 0.639 \approx 3.38\,\text{m} \]

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Key Concepts

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

Horizontal Motion
Horizontal motion refers to movement along a straight line parallel to the horizon. In projectile motion, it's often examined separately from vertical motion for clarity. In our scenario, the criminal is moving horizontally across rooftops. That's the horizontal motion component: constant and predictable.
  • Constant Speed: Here, the speed remains 5.3 m/s as no forces act horizontally (ignoring air resistance).
  • Uniform Motion: The motion is uniform, which means the horizontal velocity doesn't change over time.

It's like sliding on ice; you'll go straight unless an external force changes your path.

Vertical Motion
Vertical motion involves movement along the vertical axis, affected significantly by gravity. In this problem, the criminal's vertical motion begins once they leave the rooftop. This type of motion is characterized by acceleration due to gravity.
  • Direction: It points downward, towards the Earth.
  • Acceleration: This motion isn't constant; it's accelerating at 9.8 m/s虏 due to gravity.

Imagine dropping a ball; it starts slow but speeds up as it falls.

Uniform Motion
Uniform motion is when an object moves at a constant speed along a straight path without acceleration. It鈥檚 an idealized concept but useful.
  • Predictability: The path and speed remain unchanged, making calculations straightforward.
  • In This Scenario: Horizontally, our criminal moves uniformly after leaving the rooftop, showing a perfect application of this principle.

Think of it as cruising on a highway with cruise control on; your speed stays steady.

Gravity
Gravity is the force pulling objects toward Earth's center, crucial in projectile motion. It鈥檚 the reason the criminal doesn't float off into space.
  • Effect: Causes objects to accelerate at 9.8 m/s虏 downwards.
  • Role in the Problem: It ensures that once airborne, the criminal falls in a curved path, determining the distance and time they'll stay in flight.

Without gravity, the fall to the adjacent building wouldn鈥檛 happen!

Initial Velocity
Initial velocity is the object's speed at the very start of its motion. In our case, when the criminal jumps, this key figure helps determine their path.
  • Horizontal Aspect: Here it starts at 5.3 m/s, representing their running speed.
  • Vertical Aspect: Initially zero, as the jump is horizontal to start.

It's like the speed of a car as it pulls out of a driveway; that first push sets the stage for the journey.

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Most popular questions from this chapter

A diver springs upward from a diving board. At the instant she contacts the water, her speed is \(8.90 \mathrm{m} / \mathrm{s},\) and her body is extended at an angle of \(75.0^{\circ}\) with respect to the horizontal surface of the water. At this instant her vertical displacement is \(-3.00 \mathrm{m},\) where downward is the negative direction. Determine her initial velocity, both magnitude and direction.

A skateboarder shoots off a ramp with a velocity of \(6.6 \mathrm{m} / \mathrm{s},\) directed at an angle of \(58^{\circ}\) above the horizontal. The end of the ramp is \(1.2 \mathrm{m}\) above the ground. Let the \(x\) axis be parallel to the ground, the \(+y\) direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp. (a) How high above the ground is the highest point that the skateboarder reaches? (b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?

A major-league pitcher can throw a baseball in excess of \(41.0 \mathrm{m} / \mathrm{s}\). If a ball is thrown horizontally at this speed, how much will it drop by the time it reaches a catcher who is \(17.0 \mathrm{m}\) away from the point of release?

A spider crawling across a table leaps onto a magazine blocking its path. The initial velocity of the spider is \(0.870 \mathrm{m} / \mathrm{s}\) at an angle of \(35.0^{\circ}\) above the table, and it lands on the magazine \(0.0770 \mathrm{s}\) after leaving the table. Ignore air resistance. How thick is the magazine? Express your answer in millimeters.

A projectile is launched from ground level at an angle of \(12.0^{\circ}\) above the horizontal. It returns to ground level. To what value should the launch angle be adjusted, without changing the launch speed, so that the range doubles?
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