Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Consider a projectile launched with velocity \(\left(v_{x 0}, v_{y_{0}}\right)\) from horizontal ground (with \(x\) measured horizontally and \(y\) vertically up). Assuming no air resistance, find how long the projectile is in the air and show that the distance it travels before landing (the horizontal range) is \(2 v_{x_{0}} v_{y_{0}} / g.\)

Short Answer

Expert verified
The projectile is in the air for \(\frac{2v_{y0}}{g}\) seconds, and the horizontal range is \(\frac{2v_{x0}v_{y0}}{g}\).

Step by step solution

01

Understand the Variables

The given initial velocity is \(v_{x0}, v_{y0}\), where \(v_{x0}\) is the horizontal component and \(v_{y0}\) is the vertical component. The acceleration due to gravity is \(g\), acting downward with magnitude \(9.81 \, m/s^2\). The projectile is launched from the ground, so that the initial position is \(y = 0\).
02

Calculate Time of Flight

To find the time the projectile is in the air, use the vertical motion equation: \(y = v_{y0} t - \frac{1}{2} g t^2\). Set \(y = 0\) to solve for time, since the projectile starts and ends at ground level: \[ 0 = v_{y0} t - \frac{1}{2} g t^2 \]Factor the equation: \[ t(v_{y0} - \frac{1}{2} g t) = 0 \]This gives solutions \(t = 0\) (initial time) and \(t = \frac{2v_{y0}}{g}\). Thus, the time of flight is \(\frac{2v_{y0}}{g}\).
03

Determine the Horizontal Range

The horizontal range \(R\) is the horizontal distance the projectile travels during its flight.The horizontal motion is defined by \(x = v_{x0} t\), with \(v_{x0}\) constant because there's no horizontal acceleration.Substitute the time of flight (from Step 2): \[ R = v_{x0} \left(\frac{2v_{y0}}{g}\right) \]Simplify the expression to find the horizontal range: \[ R = \frac{2v_{x0} v_{y0}}{g} \]
04

Review the Results

We've found that the time of flight is \(\frac{2v_{y0}}{g}\) and the horizontal range is \(\frac{2v_{x0} v_{y0}}{g}\), as required by the problem statement. All calculations assume no air resistance and ideal projectile motion.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Time of Flight
The "time of flight" is a crucial concept in projectile motion, referring to the total time a projectile is in the air. To determine this, we focus on the motion in the vertical direction. The projectile is launched with an initial vertical velocity of \(v_{y0}\), and gravity acts downward with acceleration \(g = 9.81 \, \text{m/s}^2\). The vertical position of the projectile can be given by the equation:
  • \(y = v_{y0} t - \frac{1}{2} g t^2\)
Since the projectile starts and ends at the ground level, we set \(y = 0\) to solve the equation for time. This gives us two solutions: one at the initial launch \(t = 0\), and the critical time just before it hits the ground again. Solving gives \(t = \frac{2v_{y0}}{g}\), which represents the total time the projectile remains airborne.
This concept of time of flight is important because it helps predict when and where the projectile will land, allowing for strategic planning in various applications such as sports or military operations.
Horizontal Range
The "horizontal range" of a projectile is the distance it travels along the horizontal axis before landing back on the ground. To find the horizontal range, we focus on the horizontal component of motion because it remains constant (no horizontal forces if air resistance is neglected).
  • The horizontal motion is governed by: \(x = v_{x0} t\)
To find the full range of the projectile, we substitute the total time of flight from the vertical motion equation:
  • \(R = v_{x0} \cdot \frac{2v_{y0}}{g}\)
This equation highlights how both the initial horizontal and vertical velocities contribute to how far the projectile will travel.
Understanding horizontal range is particularly beneficial in fields like engineering and gaming, where the accurate prediction of motion paths is crucial.
Kinematic Equations
Kinematic equations are vital tools in physics to analyze motion, especially when it comes to projectile motion. These equations allow us to relate time, velocity, and displacement through the consideration of acceleration and initial conditions. For any projectile, we typically use:
  • \(y = v_{y0} t - \frac{1}{2} g t^2\)
  • \(x = v_{x0} t\)
In the given exercise, these equations help us separate the motion into horizontal and vertical components, manage the displacement of the projectile, and solve for time of flight and horizontal range.
By mastering kinematic equations, students can predict the outcomes of varied projectile launched scenarios. This understanding is key in designing safe and efficient trajectories in disciplines ranging from sports to aerospace engineering.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

2.16 \(\star\) A golfer hits his ball with speed \(v_{\mathrm{o}}\) at an angle \(\theta\) above the horizontal ground. Assuming that the angle \(\theta\) is fixed and that air resistance can be neglected, what is the minimum speed \(v_{\mathrm{o}}(\min )\) for which the ball will clear a wall of height \(h\), a distance \(d\) away? Your solution should get into trouble if the angle \(\tan \theta < h / d .\) Explain. What is \(v_{\mathrm{o}}(\min )\) if \(\theta=25^{\circ}, d=50 \mathrm{m},\) and \(h=2 \mathrm{m} ?\)

A gun can fire shells in any direction with the same speed \(v_{\mathrm{o}}\). Ignoring air resistance and using cylindrical polar coordinates with the gun at the origin and \(z\) measured vertically up, show that the gun can hit any object inside the surface $$z=\frac{v_{0}^{2}}{2 g}-\frac{g}{2 v_{0}^{2}} \rho^{2}$$ Describe this surface and comment on its dimensions.

The hyperbolic functions cosh \(z\) and \(\sinh z\) are defined as follows: $$\cosh z=\frac{e^{z}+e^{-z}}{2} \quad \text { and } \quad \sinh z=\frac{e^{z}-e^{-z}}{2}$$ for any \(z,\) real or complex. (a) Sketch the behavior of both functions over a suitable range of real values of \(z\). (b) Show that \(\cosh z=\cos (i z)\). What is the corresponding relation for \(\sinh z ?\) (c) What are the derivatives of cosh \(z\) and \(\sinh z ?\) What about their integrals? ( \(\mathbf{d}\) ) Show that \(\cosh ^{2} z-\sinh ^{2} z=1\) (e) Show that \(\int d x / \sqrt{1+x^{2}}=\operatorname{arcsinh} x\). [Hint: One way to do this is to make the substitution \(x=\sinh z .]\)

Consider the complex number \(z=e^{i \theta}=\cos \theta+i \sin \theta .\) (a) By evaluating \(z^{2}\) two different ways, prove the trig identities \(\cos 2 \theta=\cos ^{2} \thetaQEDQED-\sin ^{2} \theta\) and \(\sin 2 \theta=2 \sin \theta \cos \theta .\) (b) Use the same technique to find corresponding identities for \(\cos 3 \theta\) and \(\sin 3 \theta\).

A charged particle of mass \(m\) and positive charge \(q\) moves in uniform electric and magnetic fields, \(\mathbf{E}\) and \(\mathbf{B}\), both pointing in the \(z\) direction. The net force on the particle is \(\mathbf{F}=q(\mathbf{E}+\mathbf{v} \times \mathbf{B})\). Write down the equation of motion for the particle and resolve it into its three components. Solve the equations and describe the particle's motion.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free