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Ch 04: Kinematics in Two Dimensions
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 04: Kinematics in Two DimensionsProblem 78a
Chapter 4, Problem 78a

In Problems 78, 79, and 80 you are given the equations that are used to solve a problem. For each of these, you are to write a realistic problem for which these are the correct equations. Be sure that the answer your problem requests is consistent with the equations given.
100m=0m+(50cosθm/s)t10m=0m+(50sinθm/s)t1−12(9.80m/s2)t12\(\begin{aligned}\)100 \, \(\text{m}\) &= 0 \, \(\text{m}\) + (50 \(\cos\) \(\theta\) \, \(\text{m/s}\)) t_1 \\0 \, \(\text{m}\) &= 0 \, \(\text{m}\) + (50 \(\sin\) \(\theta\) \, \(\text{m/s}\)) t_1 - \(\frac{1}{2}\) (9.80 \, \(\text{m/s}\)^2) t_1^2\(\end{aligned}\)

Verified step by step guidance
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Step 1: Analyze the given equations. The first equation, 100 m = 0 m + (50cos⁡θ m/s)t₁, represents the horizontal motion of a projectile, where the horizontal displacement is 100 m, the initial horizontal velocity is (50cosθ) m/s, and t₁ is the time of flight. The second equation, 0 m = 0 m + (50sin⁡θ m/s)t₁ - (1/2)(9.80 m/s²)t₁², represents the vertical motion of the projectile, where the vertical displacement is 0 m (it returns to the same vertical level), the initial vertical velocity is (50sinθ) m/s, and the acceleration due to gravity is -9.80 m/s².
Step 2: Recognize the type of problem. These equations describe the motion of a projectile launched at an angle θ with an initial velocity of 50 m/s. The horizontal displacement is 100 m, and the projectile lands back at the same vertical level from which it was launched. The problem likely involves finding the angle θ or the time of flight t₁.
Step 3: Set up a realistic problem. A possible problem could be: 'A ball is launched from the ground with an initial velocity of 50 m/s at an angle θ above the horizontal. The ball travels a horizontal distance of 100 m before landing back on the ground. Determine the angle θ at which the ball was launched and the time of flight t₁.'
Step 4: Solve for time of flight t₁ using the horizontal motion equation. Rearrange the first equation to isolate t₁: t₁ = 100 m / (50cosθ m/s). This equation relates the time of flight to the horizontal velocity component and the horizontal displacement.
Step 5: Substitute t₁ into the vertical motion equation to solve for θ. Replace t₁ in the second equation with the expression derived in Step 4. This will result in an equation involving θ that can be solved using trigonometric identities or numerical methods. The solution will yield the angle θ that satisfies both the horizontal and vertical motion equations.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Projectile Motion

Projectile motion refers to the motion of an object that is launched into the air and is subject to the force of gravity. It can be analyzed in two dimensions: horizontal and vertical. The horizontal motion is uniform, while the vertical motion is influenced by gravitational acceleration. Understanding the components of projectile motion is essential for solving problems involving angles and distances, as seen in the given equations.
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Trigonometric Functions

Trigonometric functions, such as sine and cosine, relate the angles of a triangle to the ratios of its sides. In the context of projectile motion, these functions are used to resolve the initial velocity into horizontal and vertical components. The angle θ in the equations indicates the direction of the launch, and using trigonometric functions allows for the calculation of the object's trajectory and displacement.
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Kinematic Equations

Kinematic equations describe the motion of objects under constant acceleration. They relate displacement, initial velocity, final velocity, acceleration, and time. The equations provided in the question are specific kinematic equations that can be used to determine the position and velocity of a projectile at any given time, making them crucial for formulating realistic problems based on the motion of the object.
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Related Practice
Textbook Question

The cannon in FIGURE CP4.83 fires a projectile at launch angle θ with respect to the slope, which is at angle Φ. Find the launch angle that maximizes d. Hint: Choosing the proper coordinate system is essential. There are two options.

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Textbook Question

A cannon on a flat railroad car travels to the east with its barrel tilted 30° above horizontal. It fires a cannonball at 50 m/s. At t = 0 s , the car, starting from rest, begins to accelerate to the east at 2.0 m/s². At what time should the cannon be fired to hit a target on the tracks that is 400 m to the east of the car's initial position? Assume that the cannonball is fired from ground level.

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Textbook Question

A 6.0-cm-diameter gear rotates with angular velocity ω = ( 20 ─ ½ t² ) rad/s where t is in seconds. At t = 4.0 s, what are: The tangential acceleration of a tooth on the gear?

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Textbook Question

A painted tooth on a spinning gear has angular position θ = (6.0 rad/s⁴)t⁴. What is the tooth's angular acceleration at the end of 10 revolutions?

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Textbook Question

An archer standing on a 15° slope shoots an arrow 20° above the horizontal, as shown in FIGURE CP4.82. How far down the slope does the arrow hit if it is shot with a speed of 5.0 m/s from 1.75 m above the ground?

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Textbook Question

A 6.0-cm-diameter gear rotates with angular velocity ω = ( 20 ─ ½ t² ) rad/s where t is in seconds. At t = 4.0 s, what are: The gear's angular acceleration?