<< Chapter < Page Chapter >> Page >

Given these assumptions, the following steps are then used to analyze projectile motion:

Step 1. Resolve or break the motion into horizontal and vertical components along the x- and y-axes. These axes are perpendicular, so A x = A cos θ size 12{A rSub { size 8{x} } =A"cos"θ} {} and A y = A sin θ size 12{A rSub { size 8{y} } =A"sin"θ} {} are used. The magnitude of the components of displacement s size 12{s} {} along these axes are x size 12{x} {} and y. size 12{y} {} The magnitudes of the components of the velocity v size 12{v} {} are 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, as shown in [link] . Initial values are denoted with a subscript 0, as usual.

Step 2. Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms:

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."} {}
Vertical Motion ( assuming positive is up a y = g = 9. 80 m/s 2 ) size 12{"Vertical Motion " \( "assuming positive is up "a rSub { size 8{y} } = - g= - 9/"80"" m/s" rSup { size 8{2} } \) } {}
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 \( y - y rSub { size 8{0} } \) "."} {}

Step 3. Solve for the unknowns in the two separate motions—one horizontal and one vertical. Note that the only common variable between the motions is time t size 12{t} {} . The problem solving procedures here are the same as for one-dimensional kinematics    and are illustrated in the solved examples below.

Step 4. Recombine the two motions to find the total displacement s size 12{s} {} and velocity v size 12{v} {} . Because the x - and y -motions are perpendicular, we determine these vectors by using the techniques outlined in the Vector Addition and Subtraction: Analytical Methods and employing A = A x 2 + A y 2 size 12{v= sqrt {v rSub { size 8{x} } rSup { size 8{2} } +v rSub { size 8{y} } rSup { size 8{2} } } } {} and θ = tan 1 ( A y / A x ) size 12{θ="tan" rSup { size 8{ - 1} } \( A rSub { size 8{y} } /A rSub { size 8{x} } \) } {} in the following form, where θ size 12{θ} {} is the direction of the displacement s size 12{s} {} and θ v size 12{θ rSub { size 8{v} } } {} is the direction of the velocity v size 12{v} {} :

Total displacement and velocity

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} } \( v rSub { size 8{y} } /v rSub { size 8{x} } \) "."} {}
In part a the figure shows projectile motion of a ball with initial velocity of v zero at an angle of theta zero with the horizontal x axis. The horizontal component v x and the vertical component v y at various positions of ball in the projectile path is shown. In part b only the horizontal velocity component v sub x is shown whose magnitude is constant at various positions in the path. In part c only vertical velocity component v sub y is shown. The vertical velocity component v sub y is upwards till it reaches the maximum point and then its direction changes to downwards. In part d resultant v of horizontal velocity component v sub x and downward vertical velocity component v sub y is found which makes an angle theta with the horizontal x axis. The direction of resultant velocity v is towards south east.
(a) We analyze two-dimensional projectile motion by breaking it into two independent one-dimensional motions along the vertical and horizontal axes. (b) The horizontal motion is simple, because a x = 0 size 12{a rSub { size 8{x} } =0} {} and v x size 12{v rSub { size 8{x} } } {} is thus constant. (c) The velocity in the vertical direction begins to decrease as the object rises; at its highest point, the vertical velocity is zero. As the object falls towards the Earth again, the vertical velocity increases again in magnitude but points in the opposite direction to the initial vertical velocity. (d) The x - and y -motions are recombined to give the total velocity at any given point on the trajectory.

A fireworks projectile explodes high and away

During a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75.0º above the horizontal, as illustrated in [link] . The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes?

Strategy

Because air resistance is negligible for the unexploded shell, the analysis method outlined above can be used. The motion can be broken into horizontal and vertical motions in which a x = 0 size 12{ a rSub { size 8{x} } =0} {} and a y = g size 12{ a rSub { size 8{y} } =-g} {} . We can then define x 0 size 12{x rSub { size 8{0} } } {} and y 0 size 12{y rSub { size 8{0} } } {} to be zero and solve for the desired quantities.

Questions & Answers

Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
Jyoti Reply
I only see partial conversation and what's the question here!
Crow Reply
what about nanotechnology for water purification
RAW Reply
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
what is the stm
Brian Reply
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
LITNING Reply
what is a peer
LITNING Reply
What is meant by 'nano scale'?
LITNING Reply
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
what is Nano technology ?
Bob Reply
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
Damian Reply
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
Stoney Reply
why we need to study biomolecules, molecular biology in nanotechnology?
Adin Reply
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
Adin
why?
Adin
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
Damian Reply
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
Praveena Reply
hi
Loga
what does nano mean?
Anassong Reply
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply
Practice Key Terms 7

Get the best Algebra and trigonometry course in your pocket!





Source:  OpenStax, Physics 110 at une. OpenStax CNX. Aug 29, 2013 Download for free at http://legacy.cnx.org/content/col11566/1.1
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Physics 110 at une' conversation and receive update notifications?

Ask