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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


  • 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

how do you translate this in Algebraic Expressions
linda Reply
why surface tension is zero at critical temperature
I think if critical temperature denote high temperature then a liquid stats boils that time the water stats to evaporate so some moles of h2o to up and due to high temp the bonding break they have low density so it can be a reason
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
so some one know about replacing silicon atom with phosphorous in semiconductors device?
s. Reply
how to fabricate graphene ink ?
for screen printed electrodes ?
What is lattice structure?
s. Reply
of graphene you mean?
or in general
in general
Graphene has a hexagonal structure
On having this app for quite a bit time, Haven't realised there's a chat room in it.
what is biological synthesis of nanoparticles
Sanket Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
many many of nanotubes
what is the k.e before it land
what is the function of carbon nanotubes?
I'm interested in nanotube
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
what is system testing
what is the application of nanotechnology?
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
anybody can imagine what will be happen after 100 years from now in nano tech world
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
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
silver nanoparticles could handle the job?
not now but maybe in future only AgNP maybe any other nanomaterials
I'm interested in Nanotube
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
can nanotechnology change the direction of the face of the world
Prasenjit Reply
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.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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