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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

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Brian Reply
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Rafiq
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How we are making nano material?
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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
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what king of growth are you checking .?
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why we need to study biomolecules, molecular biology in nanotechnology?
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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.
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anyone know any internet site where one can find nanotechnology papers?
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research.net
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sciencedirect big data base
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Introduction about quantum dots in nanotechnology
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absolutely yes
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there is no specific books for beginners but there is book called principle of nanotechnology
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how can I make nanorobot?
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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
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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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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