3.5 Projectile motion  (Page 6/16)

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$R=\frac{{v}_{0}^{2}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}{2\theta }_{0}}{g}\text{,}$

where ${v}_{0}$ is the initial speed and ${\theta }_{0}$ is the initial angle relative to the horizontal. The proof of this equation is left as an end-of-chapter problem (hints are given), but it does fit the major features of projectile range as described.

When we speak of the range of a projectile on level ground, we assume that $R$ is very small compared with the circumference of the Earth. If, however, the range is large, the Earth curves away below the projectile and acceleration of gravity changes direction along the path. The range is larger than predicted by the range equation given above because the projectile has farther to fall than it would on level ground. (See [link] .) If the initial speed is great enough, the projectile goes into orbit. This possibility was recognized centuries before it could be accomplished. When an object is in orbit, the Earth curves away from underneath the object at the same rate as it falls. The object thus falls continuously but never hits the surface. These and other aspects of orbital motion, such as the rotation of the Earth, will be covered analytically and in greater depth later in this text.

Once again we see that thinking about one topic, such as the range of a projectile, can lead us to others, such as the Earth orbits. In Addition of Velocities , we will examine the addition of velocities, which is another important aspect of two-dimensional kinematics and will also yield insights beyond the immediate topic.

Phet explorations: projectile motion

Blast a Buick out of a cannon! Learn about projectile motion by firing various objects. Set the angle, initial speed, and mass. Add air resistance. Make a game out of this simulation by trying to hit a target.

Summary

• Projectile motion is the motion of an object through the air that is subject only to the acceleration of gravity.
• To solve projectile motion problems, perform the following steps:
1. Determine a coordinate system. Then, resolve the position and/or velocity of the object in the horizontal and vertical components. The components of position $\mathbf{s}$ are given by the quantities $x$ and $y$ , and the components of the velocity $\mathbf{v}$ are given by ${v}_{x}=v\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta$ and ${v}_{y}=v\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta$ , where $v$ is the magnitude of the velocity and $\theta$ is its direction.
2. Analyze the motion of the projectile in the horizontal direction using the following equations:
$\text{Horizontal motion}\left({a}_{x}=0\right)$
$x={x}_{0}+{v}_{x}t$
${v}_{x}={v}_{0x}={\mathbf{\text{v}}}_{\text{x}}=\text{velocity is a constant.}$
3. Analyze the motion of the projectile in the vertical direction using the following equations:
$\text{Vertical motion}\left(\text{Assuming positive direction is up;}\phantom{\rule{0.25em}{0ex}}{a}_{y}=-g=-9\text{.}\text{80 m}{\text{/s}}^{2}\right)$
$y={y}_{0}+\frac{1}{2}\left({v}_{0y}+{v}_{y}\right)t$
${v}_{y}={v}_{0y}-\text{gt}$
$y={y}_{0}+{v}_{0y}t-\frac{1}{2}{\text{gt}}^{2}$
${v}_{y}^{2}={v}_{0y}^{2}-2g\left(y-{y}_{0}\right).$
4. Recombine the horizontal and vertical components of location and/or velocity using the following equations:
$s=\sqrt{{x}^{2}+{y}^{2}}$
$\theta ={\text{tan}}^{-1}\left(y/x\right)$
$v=\sqrt{{v}_{x}^{2}+{v}_{y}^{2}}$
${\theta }_{\text{v}}={\text{tan}}^{-1}\left({v}_{y}/{v}_{x}\right).$
• The maximum height $h$ of a projectile launched with initial vertical velocity ${v}_{0y}$ is given by
$h=\frac{{v}_{0y}^{2}}{2g}.$
• The maximum horizontal distance traveled by a projectile is called the range . The range $R$ of a projectile on level ground launched at an angle ${\theta }_{0}$ above the horizontal with initial speed ${v}_{0}$ is given by
$R=\frac{{v}_{0}^{2}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}{2\theta }_{0}}{g}.$

can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
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Tamia
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a perfect square v²+2v+_
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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J, combine like terms 7x-4y
im not good at math so would this help me
yes
Asali
I'm not good at math so would you help me
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what is the problem that i will help you to self with?
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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what's the easiest and fastest way to the synthesize AgNP?
China
Cied
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I start with an easy one. carbon nanotubes woven into a long filament like a string
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many many of nanotubes
Porter
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what is the function of carbon nanotubes?
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In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
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after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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