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A projectile's launch speed is five times its speed at maximum height. Find launch angle \(\theta_{0}\).

Short Answer

Expert verified
The launch angle \(\theta_0\) is \(\cos^{-1}(1/5)\).

Step by step solution

01

Understanding the Relationship Between Launch Speed and Maximum Height Speed

The launch speed of the projectile is denoted as \(v_0\), and at maximum height, the vertical component of the speed is zero while the horizontal component remains constant. This implies the maximum height speed is \(v_0 \cos \theta_0\). Given that the launch speed \(v_0 = 5 \, \times\) the speed at maximum height, we have:\[v_0 = 5 \, (v_0 \cos \theta_0) \]This equation must be solved to find \(\theta_0\).
02

Simplifying the Equation

From the previous step, we have the equation:\[v_0 = 5 \, (v_0 \cos \theta_0) \]We can simplify this by dividing both sides by \(v_0\) (assuming \(v_0 eq 0\)), which gives us:\[1 = 5 \cos \theta_0\]
03

Solving for the Launch Angle \(\theta_0\)

The equation from the previous step, \(1 = 5 \cos \theta_0\), can be rearranged to find \(\cos \theta_0\):\[\cos \theta_0 = \frac{1}{5}\]Using the inverse cosine function, we can solve for \(\theta_0\):\[\theta_0 = \cos^{-1}\left(\frac{1}{5}\right)\]

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Key Concepts

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

Launch Speed
Launch speed is a fundamental concept in projectile motion, referring to the initial velocity at which an object is projected into the air. This speed is crucial because it affects the distance and height the projectile can achieve. To understand launch speed, it's key to consider that it is the combination of all velocities a projectile has when it is launched.
Most importantly, a projectile's motion can be broken down into two components: the horizontal and vertical components. The launch speed is the combination of these two and can be calculated using the Pythagorean theorem: \( v_0 = \sqrt{v_{0x}^2 + v_{0y}^2} \). Here, \( v_{0x} \) is the horizontal component, and \( v_{0y} \) is the vertical component of the speed.
Remember that launch speed stays constant in an ideal scenario without air resistance. Understanding this concept helps determine how far and high the projectile will travel.
Maximum Height
The maximum height of a projectile is the highest point it reaches in its trajectory. At this point, the vertical component of the projectile's velocity is zero. This happens because the force of gravity, acting downward, has decelerated the upward motion of the projectile until it momentarily stops before beginning to fall back to the ground.
The equation for maximum height is derived from the principle of energy conservation or by using kinematics. It can be expressed as: \( h_{max} = \frac{v_{0y}^2}{2g} \), where \( v_{0y} \) is the initial vertical component of the launch speed, and \( g \) is the acceleration due to gravity (approximately \(9.8 m/s^2\) on Earth).
Understanding maximum height is important for solving problems involving projectile motion, such as determining fields for sporting events, launching vehicles, or predicting flight paths in physics applications.
Launch Angle
The launch angle is the angle at which a projectile is fired with respect to the horizontal plane. It plays a significant role in determining the range, flight path, and height of the projectile. The launch angle can be visualized as the tilt between the direction of the launch and the ground.
To find the launch angle in projectile motion, the relationship \( \theta_0 = \cos^{-1}\left( \frac{1}{5} \right) \) might be used if given specific conditions, such as in the problem where the launch speed is five times the speed at maximum height.
Mathematically, the components of the launch speed can be expressed as \( v_{0x} = v_0 \cos \theta_0 \) and \( v_{0y} = v_0 \sin \theta_0 \). Each component affects different aspects of the projectile's motion, such as height, time in the air, and distance traveled.
In summary, the launch angle is vital for calculating the trajectory of a projectile, and knowing how to manipulate it can help in achieving the desired results in practical applications.
Horizontal Component
The horizontal component of a projectile's velocity is a measure of how fast it is moving along the horizontal axis relative to the ground. Unlike the vertical component, which changes due to gravity, the horizontal component typically remains constant if air resistance is negligible.
To find the horizontal component of a projectile's motion, the equation \( v_{0x} = v_0 \cos \theta_0 \) is used, where \( \theta_0 \) is the launch angle, and \( v_0 \) is the launch speed. This component determines how far away from the launch point the projectile will land. It basically influences the range of the projectile.
In most real-world scenarios, understanding the horizontal component helps in designing projectiles for maximum distance, such as in sports or engineering tasks. By optimizing this component, one can maximize the range while maintaining the initial launch conditions.

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Most popular questions from this chapter

You are kidnapped by political-science majors (who are upset because you told them political science is not a real science). Although blindfolded, you can tell the speed of their car (by the whine of the engine), the time of travel (by mentally counting off seconds), and the direction of travel (by turns along the rectangular street system). From these clues, you know that you are taken along the following course: \(50 \mathrm{~km} / \mathrm{h}\) for \(2.0 \mathrm{~min}\), turn \(90^{\circ}\) to the right, \(20 \mathrm{~km} / \mathrm{h}\) for \(4.0 \mathrm{~min}\), turn \(90^{\circ}\) to the right, \(20 \mathrm{~km} / \mathrm{h}\) for \(60 \mathrm{~s}\), turn \(90^{\circ}\) to the left, \(50 \mathrm{~km} / \mathrm{h}\) for \(60 \mathrm{~s}\), turn \(90^{\circ}\) to the right, \(20 \mathrm{~km} / \mathrm{h}\) for \(2.0 \mathrm{~min}\), turn \(90^{\circ}\) to the left, \(50 \mathrm{~km} / \mathrm{h}\) for \(30 \mathrm{~s}\). At that point, (a) how far are you from your starting point, and (b) in what direction relative to your initial direction of travel are you?

The position vector \(\vec{r}\) of a particle moving in the \(x y\) plane is \(\vec{r}=2 \hat{i}+2 \sin [(\pi / 4 \mathrm{rad} / \mathrm{s}) t] \hat{\mathrm{j}}, \quad\) with \(\vec{r}\) in meters and \(t\) in seconds. (a) Calculate the \(x\) and \(y\) components of the particle's position at \(t=0,1.0,2.0,3.0\), and \(4.0 \mathrm{~s}\) and sketch the particle's path in the \(x y\) plane for the interval \(0 \leq t \leq\) \(4.0 \mathrm{~s}\). (b) Calculate the components of the particle's velocity at \(t=1.0,2.0\), and \(3.0 \mathrm{~s}\). Show that the velocity is tangent to the path of the particle and in the direction the particle is moving at each time by drawing the velocity vectors on the plot of the particle's path in part (a). (c) Calculate the components of the particle's acceleration at \(t=1.0,2.0\), and \(3.0 \mathrm{~s}\).

A centripetal-acceleration addict rides in uniform circular motion with radius \(r=3.00 \mathrm{~m}\). At one instant his acceleration is \(\vec{a}=\left(6.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(-4.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\). At that instant, what are the val- ues of (a) \(\vec{v} \cdot \vec{a}\) and (b) \(\vec{r} \times \vec{a}\) ?

A football kicker can give the ball an initial speed of \(25 \mathrm{~m} / \mathrm{s}\). What are the (a) least and (b) greatest elevation angles at which he can kick the ball to score a field goal from a point \(50 \mathrm{~m}\) in front of goalposts whose horizontal bar is \(3.44 \mathrm{~m}\) above the ground?

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