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Chapter 2: Problem 18

A rock is tossed straight up with a speed of 20 m/s. When it returns, it falls into a hole \(10 \mathrm{m}\) deep. a. What is the rock's velocity as it hits the bottom of the hole? b. How long is the rock in the air, from the instant it is released until it hits the bottom of the hole?

Short Answer

Expert verified
a. The velocity at the bottom of the hole is found by calculating the square root of twice the distance multiplied by the gravitational constant. b. The total time the rock was in the air is the sum of the time taken to reach the maximum height and the time taken to fall into the hole.

Step by step solution

01

Calculate time until the rock reaches its maximum height

Using the kinematics formula \(v_f = v_i - gt\), where \(v_f\) is the final speed (which is 0 at maximum height), \(v_i\) is the initial speed (20 m/s), \(g\) is acceleration due to gravity (\(9.81 \mathrm{m/s}^2\)), and \(t\) is time, we solve for \(t\). Hence \(t = v_i / g = 20 \mathrm{m/s} / 9.81 \mathrm{m/s}^2\).
02

Calculate total distance the rock travels

The rock travels upwards until it reaches maximum height and then travels downwards into the hole. The total distance the rock travels is the sum of the distance it travels upwards and the distance it falls into the hole. This distance can be calculated from the formula \(d = v_i t + 0.5gt^2\), where \(d\) is the distance, \(v_i\) is the initial speed, \(g\) is acceleration due to gravity, and \(t\) is time. The time here is the time calculated in step 1. Hence the total distance is \(d = 20 \mathrm{m/s} * t + 0.5 * 9.81 \mathrm{m/s}^2 * t^2 + 10 \mathrm{m}\).
03

Calculate final velocity as it hits the bottom of the hole

In this step we use the formula \(v_f^2 = v_i^2 + 2gd\), where \(v_f\) is the final speed (the speed as it hits the bottom of the hole), \(v_i\) is the initial speed, \(g\) is the acceleration due to gravity, and \(d\) is the total distance traveled. Here we insert \(v_i = 0\), because the rock starts to fall from its highest point, hence the initial velocity here is zero. The distance for this part is the sum of the distance up and the distance down, i.e., the distance calculated in step 2, hence \(v_f = \sqrt{2gd}\).
04

Calculate total time the rock was in the air

The total time the rock is in the air is the sum of the time it takes for the rock to reach maximum height and the time it takes for the rock to fall into the hole. The time it takes for the rock to fall into the hole can be calculated from the formula \(t = \sqrt{2d/g}\) where \(d\) is the total distance traveled and \(g\) is the acceleration due to gravity. Hence the total time is the sum of the time calculated in step 1 and the time calculated in this step.

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Key Concepts

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

Kinematics
Kinematics is a fundamental part of mechanics that describes the motion of objects. It does not consider the forces that cause this motion, making it primarily about the "how" rather than the "why" behind movement. In our exercise, the rock's path from start to stop is an example of projectile motion in kinematics. Here are some key ideas to understand in kinematics when dealing with vertical motion:
  • Initial Velocity: This is the speed at which an object starts its journey. For example, the rock is initially tossed upwards with a velocity of 20 m/s.
  • Maximum Height: When the rock reaches the top of its path, its velocity is momentarily zero before it starts descending.
  • Displacement: This is the overall change in position of the rock as it moves upwards and downwards, ultimately landing in the hole.
Understanding these elements helps break down complex motions into simpler sequences, making it easier to solve for unknowns.
Acceleration due to Gravity
Gravity is the force that pulls objects towards the Earth. In physics, the acceleration due to gravity is denoted by the symbol \(g\) and is valued at approximately \(9.81 \text{m/s}^2\) on the surface of the Earth. It acts downwards, toward the center of the Earth, influencing objects in free fall or projectile motion.

In any scenario involving vertical motion, gravity plays a crucial role:
  • It decelerates objects as they rise. For example, the rock's upward speed will decrease at a rate of \(9.81 \text{m/s}^2\) until it stops momentarily at the maximum height.
  • It accelerates objects as they descend, adding speed at the same rate of \(9.81 \text{m/s}^2\).
This constant acceleration is vital for calculating the time taken to stop, fall, or the final velocity when hitting the ground or hole as in our exercise.
Velocity Calculation
Velocity is a vector quantity, meaning it has both magnitude and direction. Calculating velocity requires careful consideration of both the initial conditions and the forces acting on the object. In our example, the kinematics equation \(v_f = v_i - gt\) helps us find the velocity at various points in the rock's journey, with specific attention to:
  • Initial Speed: The speed at which the rock is tossed upwards, impactfully increasing its ascent velocity.
  • Final Speed: The velocity as the rock hits the bottom of the hole, which results from combining the downward gravitational acceleration from the peak of its trajectory.
The formula \(v_f^2 = v_i^2 + 2gd\) is also useful when directly calculating final velocity using the total distance traveled and the influence of gravity, especially if the initial upward motion is fully retraced on descent.
Time of Flight
Time of flight in projectile motion refers to the total duration an object remains in the air, from the time it is launched to when it returns to the surface. In the exercise about the rock, calculating this helps us know how long the rock has been airborne before landing in the hole.

The time of flight can be calculated by breaking it down into two segments:
  • The time to reach maximum height: Derived using the kinematics formula, \(t = \frac{v_i}{g}\), where the final velocity at the peak is zero.
  • The time to descend from maximum height to the bottom of the hole: Found using \(t = \sqrt{\frac{2d}{g}}\) where \(d\) is the sum of the maximum height and the distance fallen into the hole.
By adding both durations, we get the total time the rock remains in the air, giving a full picture of its motion trajectory.

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Most popular questions from this chapter

When a 1984 Alfa Romeo Spider sports car accelerates at the maximum possible rate, its motion during the first 20 s is extremely well modeled by the simple equation $$v_{x}^{2}=\frac{2 P}{m} t$$ where \(P=3.6 \times 10^{4}\) watts is the car's power output, \(m=\) \(1200 \mathrm{kg}\) is its mass, and \(v_{x}\) is in \(\mathrm{m} / \mathrm{s} .\) That is, the square of the car's velocity increases linearly with time. a. What is the car's speed at \(t=10 \mathrm{s}\) and at \(t=20 \mathrm{s} ?\) b. Find a symbolic expression, in terms of \(P, m,\) and \(t,\) for the car's acceleration at time \(\iota\) c. Evaluate the acceleration at \(t=1 \mathrm{s}\) and \(t=10 \mathrm{s}\) d. This simple model fails for \(t\) less than about \(0.5 \mathrm{s}\). Explain how you can recognize the failure. e. Find a symbolic expression for the distance the car has traveled at time \(t\) f. One-quarter mile is \(402 \mathrm{m}\). What is the Spider's best time in a quarter-mile race? (The model's failure in the first 0.5 s has very little effect on your answer because the car travels almost no distance during that time.)

The Starship Enterprise returns from warp drive to ordinary space with a forward speed of \(50 \mathrm{km} / \mathrm{s}\). To the crew's great surprise, a Klingon ship is \(100 \mathrm{km}\) directly ahead, traveling in the same direction at a mere \(20 \mathrm{km} / \mathrm{s}\). Without evasive action, the Enterprise will overtake and collide with the Klingons in just slightly over \(3.0 \mathrm{s}\). The Enterprise's computers react instantly to brake the ship. What magnitude acceleration does the Enterprise need to just barely avoid a collision with the Klingon ship? Assume the acceleration is constant. Hint: Draw a position-versus-time graph showing the motions of both the Enterprise and the Klingon ship. Let \(x_{0}=0\) km be the location of the Enterprise as it returns from warp drive. How do you show graphically the situation in which the collision is "barely avoided"? Once you decide what it looks like graphically, express that situation mathematically.

You're driving down the highway late one night at \(20 \mathrm{m} / \mathrm{s}\) when a deer steps onto the road \(35 \mathrm{m}\) in front of you. Your reaction time before stepping on the brakes is \(0.50 \mathrm{s}\), and the maximum deceleration of your car is \(10 \mathrm{m} / \mathrm{s}^{2}\) a. How much distance is between you and the deer when you come to a stop? b. What is the maximum speed you could have and still not hit the deer?

A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at \(30 \mathrm{m} / \mathrm{s}^{2}\) for \(30 \mathrm{s}\), then runs out of fuel. Ignore any air resistance effects. a. What is the rocket's maximum altitude? b. How long is the rocket in the air before hitting the ground? c. Draw a velocity-versus-time graph for the rocket from liftoff until it hits the ground.

David is driving a steady \(30 \mathrm{m} / \mathrm{s}\) when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady \(2.0 \mathrm{m} / \mathrm{s}^{2}\) at the instant when David passes. a. How far does Tina drive before passing David? b. What is her speed as she passes him?
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