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Solution for (a)

By “height” we mean the altitude or vertical position y size 12{y} {} above the starting point. The highest point in any trajectory, called the apex, is reached when v y = 0 size 12{ v rSub { size 8{y} } =0} {} . Since we know the initial and final velocities as well as the initial position, we use the following equation to find y size 12{y} {} :

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 \( y - y rSub { size 8{0} } \) "."} {}
The x y graph shows the trajectory of fireworks shell. The initial velocity of the shell v zero is at angle theta zero equal to seventy five degrees with the horizontal x axis. The fuse is set to explode the shell at the highest point of the trajectory which is at a height h equal to two hundred thirty three meters and at a horizontal distance x equal to one hundred twenty five meters from the origin.
The trajectory of a fireworks shell. The fuse is set to explode the shell at the highest point in its trajectory, which is found to be at a height of 233 m and 125 m away horizontally.

Because y 0 size 12{y rSub { size 8{0} } } {} and v y size 12{v rSub { size 8{y} } } {} are both zero, the equation simplifies to

0 = v 0 y 2 2 gy. size 12{0=v rSub { size 8{0y} } rSup { size 8{2} } - 2 ital "gy."} {}

Solving for y size 12{y} {} gives

y = v 0 y 2 2 g . size 12{y= { {v rSub { size 8{0y} } rSup { size 8{2} } } over {2g} } "." } {}

Now we must find v 0 y size 12{v rSub { size 8{0y} } } {} , the component of the initial velocity in the y -direction. It is given by v 0 y = v 0 sin θ size 12{v rSub { size 8{0y rSup} =v rSub {0 rSup size 12{"sin"θ}} {} , where v 0 y is the initial velocity of 70.0 m/s, and θ 0 = 75.0º size 12{θ rSub { size 8{0} } } {} is the initial angle. Thus,

v 0 y = v 0 sin θ 0 = ( 70.0 m/s ) ( sin 75º ) = 67.6 m/s. size 12{v rSub { size 8{0y} } =v rSub { size 8{0} } "sin"θ rSub { size 8{0} } = \( "70" "." 0" m/s" \) \( "sin""75" { size 12{ circ } } \) ="67" "." 6" m/s."} {}

and y size 12{y} {} is

y = ( 67 .6 m/s ) 2 2 ( 9 . 80 m /s 2 ) , size 12{y= { { \( "67" "." 6" m/s" \) rSup { size 8{2} } } over {2 \( 9 "." "80"" m/s" rSup { size 8{2} } \) } } } {}

so that

y = 233 m. size 12{y="233"" m."} {}

Discussion for (a)

Note that because up is positive, the initial velocity is positive, as is the maximum height, but the acceleration due to gravity is negative. Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6 m/s initial vertical component of velocity will reach a maximum height of 233 m (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding. In practice, air resistance is not completely negligible, and so the initial velocity would have to be somewhat larger than that given to reach the same height.

Solution for (b)

As in many physics problems, there is more than one way to solve for the time to the highest point. In this case, the easiest method is to use 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} {} . Because y 0 size 12{y rSub { size 8{0} } } {} is zero, this equation reduces to simply

y = 1 2 ( v 0 y + v y ) t . size 12{y= { {1} over {2} } \( v rSub { size 8{0y} } +v rSub { size 8{y} } \) t} {}

Note that the final vertical velocity, v y size 12{v rSub { size 8{y} } } {} , at the highest point is zero. Thus,

t = 2 y ( v 0y + v y ) = 2 ( 233 m ) ( 67.6 m/s ) = 6.90 s . alignl { stack { size 12{t= { {2y} over { \( v rSub { size 8{0y} } +v rSub { size 8{y} } \) } } = { {2 times "233"" m"} over { \( "67" "." 6" m/s" \) } } } {} #=6 "." "90"" s" {} } } {}

Discussion for (b)

This time is also reasonable for large fireworks. When you are able to see the launch of fireworks, you will notice several seconds pass before the shell explodes. (Another way of finding the time is by using 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} } } {} , and solving the quadratic equation for t size 12{t} {} .)

Solution for (c)

Because air resistance is negligible, a x = 0 size 12{a rSub { size 8{x} } =0} {} and the horizontal velocity is constant, as discussed above. The horizontal displacement is horizontal velocity multiplied by time as given by x = x 0 + v x t size 12{x=x rSub { size 8{0} } +v rSub { size 8{x} } t} {} , where x 0 size 12{x rSub { size 8{0} } } {} is equal to zero:

x = v x t , size 12{x=v rSub { size 8{x} } t ","} {}

where v x size 12{v rSub { size 8{x} } } {} is the x -component of the velocity, which is given by v x = v 0 cos θ 0 . size 12{v rSub { size 8{x} } =v rSub { size 8{0} } "cos"θ rSub { size 8{0} } "." } {} Now,

v x = v 0 cos θ 0 = ( 70 . 0 m/s ) ( cos 75.0º ) = 18 . 1 m/s. size 12{v rSub { size 8{x} } =v rSub { size 8{0} } "cos"θ rSub { size 12{0} } = \( "70" "." 0" m/s" \) \( "cos""75.0º" \) ="18" "." 1" m/s."} {}

The time t size 12{t} {} for both motions is the same, and so x size 12{t} {} is

x = ( 18 . 1 m/s ) ( 6 . 90 s ) = 125 m. size 12{x= \( "18" "." 1" m/s" \) \( 6 "." "90"" s" \) ="125"" m."} {}

Discussion for (c)

The horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. Once the shell explodes, air resistance has a major effect, and many fragments will land directly below.

In solving part (a) of the preceding example, the expression we found for y size 12{y} {} is valid for any projectile motion where air resistance is negligible. Call the maximum height y = h size 12{y=h} {} ; then,

Questions & Answers

where we get a research paper on Nano chemistry....?
Maira Reply
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
Maira Reply
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
Google
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
Hafiz Reply
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
Jyoti Reply
ya I also want to know the raman spectra
Bhagvanji
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
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
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
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
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
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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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