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R = v 0 2 sin 2 θ 0 g , size 12{R= { {v rSub { size 8{0} } rSup { size 8{2} } "sin"2θ rSub { size 8{0} } } over {g} } ","} {}

where v 0 size 12{v rSub { size 8{0} } } {} is the initial speed and θ 0 size 12{θ rSub { size 8{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 size 12{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.

A figure of the Earth is shown and on top of it a very high tower is placed. A projectile satellite is launched from this very high tower with initial velocity of v zero in the horizontal direction. Several trajectories are shown with increasing range. A circular trajectory is shown indicating the satellite achieved its orbit and it is revolving around the Earth.
Projectile to satellite. In each case shown here, a projectile is launched from a very high tower to avoid air resistance. With increasing initial speed, the range increases and becomes longer than it would be on level ground because the Earth curves away underneath its path. With a large enough initial speed, orbit is achieved.

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.

Projectile Motion

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 s size 12{s} {} are given by the quantities x size 12{x} {} and y size 12{y} {} , and the components of the velocity v size 12{v} {} are given by v x = v cos θ size 12{v rSub { size 8{x} } =v"cos"θ} {} and v y = v sin θ size 12{v rSub { size 8{y} } =v"sin"θ} {} , where v size 12{v} {} is the magnitude of the velocity and θ size 12{θ} {} is its direction.
    2. Analyze the motion of the projectile in the horizontal direction using the following equations:
      Horizontal motion ( a x = 0 ) size 12{"Horizontal motion " \( a rSub { size 8{x} } =0 \) } {}
      x = x 0 + v x t size 12{x=x rSub { size 8{0} } +v rSub { size 8{x} } t} {}
      v x = v 0 x = v x = velocity is a constant. size 12{v rSub { size 8{x} } =v rSub { size 8{0x} } =v rSub { size 8{x} } ="velocity is a constant."} {}
    3. Analyze the motion of the projectile in the vertical direction using the following equations:
      Vertical motion ( Assuming positive direction is up; a y = g = 9 . 80 m /s 2 ) size 12{"Vertical motion " \( "Assuming positive direction is up; "a rSub { size 8{y} } = - g= - 9 "." "80"" m/s" rSup { size 8{2} } \) } {}
      y = y 0 + 1 2 ( v 0 y + v y ) t size 12{y=y rSub { size 8{0} } + { {1} over {2} } \( v rSub { size 8{0y} } +v rSub { size 8{y} } \) t} {}
      v y = v 0 y gt size 12{v rSub { size 8{y} } =v rSub { size 8{0y} } - ital "gt"} {}
      y = y 0 + v 0 y t 1 2 gt 2 size 12{y=y rSub { size 8{0} } +v rSub { size 8{0y} } t - { {1} over {2} } ital "gt" rSup { size 8{2} } } {}
      v y 2 = v 0 y 2 2 g ( y y 0 ) . size 12{v rSub { size 8{y} } rSup { size 8{2} } =v rSub { size 8{0y} } rSup { size 8{2} } - 2g \( y - y rSub { size 8{0} } \) } {}
    4. Recombine the horizontal and vertical components of location and/or velocity using the following equations:
      s = x 2 + y 2 size 12{s= sqrt {x rSup { size 8{2} } +y rSup { size 8{2} } } } {}
      θ = tan 1 ( y / x ) size 12{θ="tan" rSup { size 8{ - 1} } \( y/x \) } {}
      v = v x 2 + v y 2 size 12{v= sqrt {v rSub { size 8{x} } rSup { size 8{2} } +v rSub { size 8{y} } rSup { size 8{2} } } } {}
      θ v = tan 1 ( v y / v x ) . size 12{θ rSub { size 8{v} } ="tan" rSup { size 8{ - 1} } \( v rSub { size 8{y} } /v rSub { size 8{x} } \) } {}
  • The maximum height h size 12{h} {} of a projectile launched with initial vertical velocity v 0 y size 12{v rSub { size 8{0y} } } {} is given by
    h = v 0 y 2 2 g . size 12{h= { {v rSub { size 8{0y} } rSup { size 8{2} } } over {2g} } } {}
  • The maximum horizontal distance traveled by a projectile is called the range . The range R size 12{R} {} of a projectile on level ground launched at an angle θ 0 size 12{θ rSub { size 8{0} } } {} above the horizontal with initial speed v 0 size 12{v rSub { size 8{0} } } {} is given by
    R = v 0 2 sin 2 θ 0 g . size 12{R= { {v rSub { size 8{0} } rSup { size 8{2} } "sin"2θ rSub { size 8{0} } } over {g} } } {}

Questions & Answers

what does nano mean?
Anassong Reply
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
Damian Reply
absolutely yes
Daniel
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it is a goid question and i want to know the answer as well
Maciej
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Anassong
Do somebody tell me a best nano engineering book for beginners?
s. Reply
what is fullerene does it is used to make bukky balls
Devang Reply
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Abhijith Reply
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
so some one know about replacing silicon atom with phosphorous in semiconductors device?
s. Reply
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
SUYASH Reply
for screen printed electrodes ?
SUYASH
What is lattice structure?
s. Reply
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
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Sanket Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
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what is the function of carbon nanotubes?
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what is system testing?
AMJAD
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how to synthesize TiO2 nanoparticles by chemical methods
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how did you get the value of 2000N.What calculations are needed to arrive at it
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Source:  OpenStax, Physics 110 at une. OpenStax CNX. Aug 29, 2013 Download for free at http://legacy.cnx.org/content/col11566/1.1
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