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Chapter 3: Problem 45

Two cannons are mounted as shown in the drawing and rigged to fire simultaneously. They are used in a circus act in which two clowns serve as human cannonballs. The clowns are fired toward each other and collide at a height of \(1.00 \mathrm{~m}\) above the muzzles of the cannons. Clown A is launched at a \(75.0^{\circ}\) angle, with a speed of \(9.00 \mathrm{~m} / \mathrm{s}\). The horizontal separation between the clowns as they leave the cannons is \(6.00 \mathrm{~m} .\) Find the launch speed \(v_{0 B}\) and the launch angle \(\theta_{B}\left(>45.0^{\circ}\right)\) for clown B.

Short Answer

Expert verified
The launch speed for Clown B is determined by solving the equations derived from the horizontal and vertical motion components considering the collision conditions.

Step by step solution

01

Analyze the Motion for Clown A

Clown A is launched at an angle of \(75.0^{\circ}\) with an initial speed of \(9.00 \text{ m/s}\). Break down the initial velocity into horizontal and vertical components. Use the formulas: \(v_{0x} = v_0 \cdot \cos(\theta)\) and \(v_{0y} = v_0 \cdot \sin(\theta)\). Here, \(v_0 = 9.00 \text{ m/s}\) and \(\theta = 75.0^{\circ}\).
02

Calculate the Time of Flight for Clown A

Determine the time it takes for Clown A to reach a height of \(1.00 \text{ m}\) using the vertical motion equation: \(y = v_{0y} \cdot t - \frac{1}{2}gt^2\). Here, \(y = 1.00 \text{ m}\), \(v_{0y}\) is from Step 1, and \(g = 9.81 \text{ m/s}^2\). Solve for \(t\).
03

Use Horizontal Motion to Analyze Clown B's Launch Parameters

Given that both clowns meet at a distance of \(6.00 \text{ m}\), use the horizontal component of the motion: \(x = v_{0xA} \cdot t + v_{0xB} \cdot t = 6.00 \text{ m}\). Plug in \(v_{0xA}\) from Step 1 and \(t\) from Step 2 to find \(v_{0xB}\).
04

Determine Launch Speed \(v_{0B}\) of Clown B

Once \(v_{0xB}\) is calculated, use the relation \(v_{0xB} = v_{0B} \cdot \cos(\theta_B)\) to find \(v_{0B}\). Start by assuming \(\theta_B\) is greater than \(45.0^{\circ}\) and solve the equations.
05

Calculate Launch Angle \(\theta_B\) for Clown B

To find \(\theta_B\), rearrange the equation for the vertical component of Clown B's motion: \(1.00 \text{ m} = v_{0B} \cdot \sin(\theta_B) \cdot t - \frac{1}{2}gt^2\). Use previously determined values of \(v_{0B}\) and \(t\) to solve for \(\theta_B\).

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Key Concepts

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

Kinematics
In the world of physics, kinematics is all about understanding motion without considering the forces that cause it. It's like watching a car zoom by and figuring out its speed and direction without worrying about what engine it has. In our exercise, we explore the motion of clowns being shot from cannons - quite a spectacle! Kinematics helps us break their motion down into more manageable parts. We consider things like velocity, time, and distance to solve the problem.
  • Velocity: The speed of the clowns and the direction they travel.
  • Time: How long they are in the air, which is essential for understanding their paths.
  • Distance: Their horizontal separation and vertical movement, which tell us where they meet.
Kinematics acts as the backbone of solving our problem. With formulas that connect these elements, we can predict where and when the clowns collide, making it all a fantastic show!
Trigonometry in Physics
When discussing projectile motion, trigonometry steps in to help us break down complex motion into simpler calculations. The angles and sides of triangles created by our cannon launches are crucial to uncovering the velocity components our clowns experience.
To simplify calculations, we use:
  • The cosine function to find horizontal components of velocity: \(v_{0x} = v_0 \cdot \cos(\theta)\)
  • The sine function for vertical components: \(v_{0y} = v_0 \cdot \sin(\theta)\)
This way, we can break the clown's path into horizontal and vertical motions, making it easier to manage. Trigonometry helps us pick apart the initial speed and angle into these useful pieces, so we can solve for where they all end up. With these tools, it becomes much simpler to navigate the arena of projectile equations!
Horizontal and Vertical Components
Understanding the horizontal and vertical components of motion is like taking apart a puzzle. Each piece shows a part of how our clowns fly through the air. With the use of trigonometric functions, we can split the initial velocity into respective directions.
  • The horizontal component, \(v_{0x}\), dictates how far the clown travels across the arena. It stays constant because there鈥檚 no air resistance in our problem.
  • The vertical component, \(v_{0y}\), deals with how high the clown goes. This part is affected by gravity, pulling them back towards the ground.
In the exercise, we observed that these components allow us to calculate flight times, meet-up points, and launch speeds effectively. By keeping these parts in mind, we can predict and control the motion of both clowns, ensuring they meet mid-air!
Two-Dimensional Motion
The world of two-dimensional motion combines both horizontal and vertical movements. Our clowns don't just shoot straight up or from side to side - they fly through the air in a curved motion! This is where the power of two-dimensional analysis shines, enabling us to interpret and predict trajectories.
  • The horizontal movement helps us figure out where on the ground their paths cross.
  • The vertical movement shows us when they reach the proper height.
Projectile motion involves calculating these two dimensions simultaneously to forecast the collision point accurately. By understanding and applying two-dimensional motion principles, we grasp how clowns with different launch speeds and angles can meet perfectly above the circus grounds, creating a dazzling performance.

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Most popular questions from this chapter

Interactive LearningWare 3.1 at reviews the approach taken in problems such as this one. A bird watcher meanders through the woods, walking \(0.50 \mathrm{~km}\) due east, \(0.75 \mathrm{~km}\) due south, and \(2.15 \mathrm{~km}\) in a direction \(35.0^{\circ}\) north of west. The time required for this trip is \(2.50 \mathrm{~h}\). Determine the magnitude and direction (relative to due west) of the bird watcher's (a) displacement and (b) average velocity. Use kilometers and hours for distance and time, respectively.

A speed ramp at an airport is a moving conveyor belt on which you can either stand or walk. It is intended to get you from place to place more quickly. Suppose a speed ramp is \(120 \mathrm{~m}\) long. When you walk at a comfortable speed on the ground, you cover this distance in \(86 \mathrm{~s}\). When you walk on the speed ramp at this same comfortable speed, you cover this distance in 35 s. Determine the speed at which the speed ramp is moving relative to the ground.

Mario, a hockey player, is skating due south at a speed of \(7.0 \mathrm{~m} / \mathrm{s}\) relative to the ice. A teammate passes the puck to him. The puck has a speed of \(11.0 \mathrm{~m} / \mathrm{s}\) and is moving in a direction of \(22^{\circ}\) west of south, relative to the ice. What are the magnitude and direction (relative to due south) of the puck's velocity, as observed by Mario?

The lob in tennis is an effective tactic when your opponent is near the net. It consists of lofting the ball over his head, forcing him to move quickly away from the net (see the drawing). Suppose that you loft the ball with an initial speed of \(15.0 \mathrm{~m} / \mathrm{s},\) at an angle of \(50.0^{\circ}\) above the horizontal. At this instant your opponent is \(10.0 \mathrm{~m}\) away from the ball. He begins moving away from you 0.30 s later, hoping to reach the ball and hit it back at the moment that it is \(2.10 \mathrm{~m}\) above its launch point. With what minimum average speed must he move? (Ignore the fact that he can stretch, so that his racket can reach the ball before he does.)

The ball is \(26.9 \mathrm{~m}\) from the goalpost. The ball is kicked with an initial velocity of \(19.8 \mathrm{~m} / \mathrm{s}\) at an angle \(\theta\) above the ground. Between what two angles, \(\theta_{1}\) and \(\theta_{2}\), will the ball clear the \(2.74\) -m-high crossbar? (Hint: The following trigonometric identities may be useful: \(\sec \theta=1 /(\cos \theta)\) and \(\sec ^{2} \theta=1+\tan ^{2} \theta\)
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