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Chapter 3: Problem 22

At \(t=0\) a rock is projected from ground level with a speed of \(15.0 \mathrm{~m} / \mathrm{s}\) and at an angle of \(53.0^{\circ}\) above the horizontal. Neglect air resistance. At what two times \(t\) is the rock \(5.00 \mathrm{~m}\) above the ground? At each of these two times, what are the horizontal and vertical components of the velocity of the rock? Let \(v_{0 x}\) and \(v_{0 y}\) be in the positive \(x\) - and \(y\) -directions, respectively.

Short Answer

Expert verified
The rock will be 5m above the ground at two instances, the times for which can be calculated from the quadratic equation obtained in step 2. At these times, the horizontal component of the velocity remains the same and is equal to \(v_{0x}\). The vertical components at these two times can be different and can be calculated using the equation from the step 3.

Step by step solution

01

Decomposition of the Initial Velocity:

At time \(t=0\), decompose the initial velocity into its horizontal and vertical components using the angle of projection. They can be calculated as \(v_{0x} = v0 * cos(\theta)\) and \(v_{0y} = v0 * sin(\theta)\) where v0 is the initial velocity and \(\theta\) is the angle of projection. By substituting \(v0 = 15.0m/s\) and \(\theta = 53.0 ^{\circ}\), we find the horizontal and vertical components of the initial velocity.
02

Determine When the Rock Reaches 5m

Use the kinematic equation \(y = v_{0y} * t - 0.5 * g * t^2\) where y is the height, g is the acceleration due to gravity, and t is the time. We know the rock is 5m above the ground at some time, so we can set \(y = 5.0m\) and solve the equation for two values of t. The equation will be quadratic in nature, so we will have two times - one on the way up and the other on the way down.
03

Determine the Horizontal and Vertical Components of Velocity

The horizontal component of velocity \(v_x\) remains constant throughout the journey, so it should be equal to \(v_{0x}\) we calculated in step 1. For the vertical component of velocity \(v_y\), use the equation \(v_y = v_{0y} - g*t\) at both instances of time obtained from step 2 to get two different values, one for each time.

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Key Concepts

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

Kinematic Equations
Kinematic equations are crucial for analyzing the motion of objects, especially in projectile motion where we deal with objects moving in two dimensions under the influence of gravity. When dealing with projectile problems, we use a set of equations to predict how the projectile moves over time from different perspectives like distance, speed, and time.

In our exercise, we use one such kinematic equation:
  • Vertical position equation: \( y = v_{0y} \cdot t - 0.5 \cdot g \cdot t^2 \)
This equation helps us to find the height of the rock at any given time, accounting for its initial vertical velocity and the influence of gravity.

By setting this formula equal to 5.0 meters, we can solve for the times when the rock is at this height, noticing that there will be two such times 鈥� once as the rock is moving upwards and once when it is coming back down.
Velocity Components
The initial velocity of the projectile can be split into two components: horizontal and vertical. This decomposition is crucial to understand how the object moves over time. Each component is calculated using trigonometric functions based on the angle of launch.

The general formulas are:
  • Horizontal component: \( v_{0x} = v0 \cdot \cos(\theta) \)
  • Vertical component: \( v_{0y} = v0 \cdot \sin(\theta) \)
For a launch speed of 15.0 m/s and an angle of 53.0掳:
  • The horizontal component indicates how fast the rock travels across the ground.
  • Meanwhile, the vertical component determines how high the rock will go.
Interestingly, while the horizontal velocity remains constant due to the absence of air resistance, the vertical velocity changes over time due to gravity's pull.
Quadratic Equations
Quadratic equations pop up whenever you have a squared term, like in our kinematic equation for vertical motion. These equations can often have two solutions, indicative of two different times or scenarios in physics.

For our problem, when we set the vertical position equation equal to 5 meters, we have:
  • \( v_{0y} \cdot t - 0.5 \cdot g \cdot t^2 = 5 \)
This is a classic quadratic equation in the form:
  • \( a \cdot t^2 + b \cdot t + c = 0 \)
Knowing that quadratic equations can have two solutions, we solve for two separate times when the projectile reaches this height:
  • One time when it is ascending, and
  • Another when it is descending.
By applying the quadratic formula given by
  • \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
we determine these specific times, making quadratic equations a key player in solving such projectile motion problems.

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Most popular questions from this chapter

A small object is projected from level ground with an initial velocity of magnitude \(16.0 \mathrm{~m} / \mathrm{s}\) and directed at an angle of \(60.0^{\circ}\) above the horizontal. (a) What is the horizontal displacement of the object when it is at its maximum height? How does your result compare to the horizontal range \(R\) of the object? (b) What is the vertical displacement of the object when its horizontal displacement is \(80.0 \%\) of its horizontal range \(R ?\) How does your result compare to the maximum height \(h_{\max }\) reached by the object? (c) For when the object has horizontal displacement \(x-x_{0}=\alpha R,\) where \(\alpha\) is a positive constant, derive an expression (in terms of \(\alpha\) ) for \(\left(y-y_{0}\right) / h_{\max }\). Your result should not depend on the initial velocity or the angle of projection. Show that your expression gives the correct result when \(\alpha=0.80,\) as is the case in part (b). Also show that your expression gives the correct result for \(\alpha=0, \alpha=0.50\), and \(\alpha=1.0\)

Crickets Chirpy and Milada jump from the top of a vertical cliff. Chirpy drops downward and reaches the ground in \(2.70 \mathrm{~s},\) while Milada jumps horizontally with an initial speed of \(95.0 \mathrm{~cm} / \mathrm{s}\). How far from the base of the cliff will Milada hit the ground? Ignore air resistance.

You are a member of a geological team in Central Africa. Your team comes upon a wide river that is flowing east. You must determine the width of the river and the current speed (the speed of the water relative to the earth). You have a small boat with an outboard motor. By measuring the time it takes to cross a pond where the water isn't flowing, you have calibrated the throttle settings to the speed of the boat in still water. You set the throttle so that the speed of the boat relative to the river is a constant \(6.00 \mathrm{~m} / \mathrm{s}\). Traveling due north across the river, you reach the opposite bank in \(20.1 \mathrm{~s}\). For the return trip, you change the throttle setting so that the speed of the boat relative to the water is \(9.00 \mathrm{~m} / \mathrm{s}\). You travel due south from one bank to the other and cross the river in \(11.2 \mathrm{~s}\). (a) How wide is the river, and what is the current speed? (b) With the throttle set so that the speed of the boat relative to the water is \(6.00 \mathrm{~m} / \mathrm{s},\) what is the shortest time in which you could cross the river, and where on the far bank would you land?

In fighting forest fires, airplanes work in support of ground crews by dropping water on the fires. For practice, a pilot drops a canister of red dye, hoping to hit a target on the ground below. If the plane is flying in a horizontal path \(90.0 \mathrm{~m}\) above the ground and has a speed of \(64.0 \mathrm{~m} / \mathrm{s}(143 \mathrm{mi} / \mathrm{h}),\) at what horizontal distance from the target should the pilot release the canister? Ignore air resistance.

The radius of the earth's orbit around the sun (assumed to be circular) is \(1.50 \times 10^{8} \mathrm{~km},\) and the earth travels around this orbit in 365 days. (a) What is the magnitude of the orbital velocity of the earth, in \(\mathrm{m} / \mathrm{s} ?(\mathrm{~b})\) What is the magnitude of the radial acceleration of the earth toward the sun, in \(\mathrm{m} / \mathrm{s}^{2} ?\) (c) Repeat parts (a) and (b) for the motion of the planet Mercury (orbit radius \(=5.79 \times 10^{7} \mathrm{~km},\) orbital period \(=88.0\) days).
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