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t = b ± b 2 4 ac 2 a . size 12{t= { { - b +- sqrt {b rSup { size 8{2} } - 4 ital "ac"} } over {"2a"} } "." } {}

This equation yields two solutions: t = 3.96 size 12{t=3 "." "96"} {} and t = 1.03 size 12{t=3 "." "96"} {} . (It is left as an exercise for the reader to verify these solutions.) The time is t = 3.96 s size 12{t=3 "." "96""s"} {} or 1.03 s size 12{-1 "." "03""s"} {} . The negative value of time implies an event before the start of motion, and so we discard it. Thus,

t = 3 . 96 s . size 12{t=3 "." "96"" s."} {}

Discussion for (a)

The time for projectile motion is completely determined by the vertical motion. So any projectile that has an initial vertical velocity of 14.3 m/s and lands 20.0 m below its starting altitude will spend 3.96 s in the air.

Solution for (b)

From the information now in hand, we can find the final horizontal and vertical velocities v x size 12{v rSub { size 8{x} } } {} and v y size 12{v rSub { size 8{y} } } {} and combine them to find the total velocity v size 12{v} {} and the angle θ 0 size 12{θ rSub { size 8{0} } } {} it makes with the horizontal. Of course, v x size 12{v rSub { size 8{x} } } {} is constant so we can solve for it at any horizontal location. In this case, we chose the starting point since we know both the initial velocity and initial angle. Therefore:

v x = v 0 cos θ 0 = ( 25 . 0 m/s ) ( cos 35º ) = 20 . 5 m/s. size 12{v rSub { size 8{x} } =v rSub { size 8{0} } "cos"θ rSub { size 8{0} } = \( "25" "." 0" m/s" \) \( "cos""35" rSup { size 8{ circ } } \) ="20" "." 5" m/s."} {}

The final vertical velocity is given by the following equation:

v y = v 0 y gt, size 12{v rSub { size 8{y} } =v rSub { size 8{0y} } - ital "gt,"} {}

where v 0y size 12{v rSub { size 8{0y} } } {} was found in part (a) to be 14 . 3 m/s size 12{"14" "." "3 m/s"} {} . Thus,

v y = 14 . 3 m/s ( 9 . 80 m/s 2 ) ( 3 . 96 s ) size 12{v rSub { size 8{y} } ="14" "." 3" m/s" - \( 9 "." "80"" m/s" rSup { size 8{2} } \) \( 3 "." "96"" s" \) } {}

so that

v y = 24 . 5 m/s. size 12{v rSub { size 8{y} } = - "24" "." 5" m/s."} {}

To find the magnitude of the final velocity v size 12{v} {} we combine its perpendicular components, using the following equation:

v = v x 2 + v y 2 = ( 20 . 5 m/s ) 2 + ( 24 . 5 m/s ) 2 , size 12{v= sqrt {v rSub { size 8{x} } rSup { size 8{2} } +v rSub { size 8{y} } rSup { size 8{2} } } = sqrt { \( "20" "." 5" m/s" \) rSup { size 8{2} } + \( - "24" "." 5" m/s" \) rSup { size 8{2} } } ","} {}

which gives

v = 31 . 9 m/s. size 12{v="31" "." 9" m/s."} {}

The direction θ v size 12{θ rSub { size 8{v} } } {} is found from the equation:

θ v = tan 1 ( v y / v x ) size 12{θ rSub { size 8{v} } ="tan" rSup { size 8{ - 1} } \( v rSub { size 8{y} } /v rSub { size 8{x} } \) } {}

so that

θ v = tan 1 ( 24 . 5 / 20 . 5 ) = tan 1 ( 1 . 19 ) . size 12{θ rSub { size 8{v} } ="tan" rSup { size 8{ - 1} } \( - "24" "." 5/"20" "." 5 \) ="tan" rSup { size 8{ - 1} } \( - 1 "." "19" \) "."} {}


θ v = 50 . 1 º . size 12{θ rSub { size 8{v} } = - "50" "." 1 rSup { size 12{ circ } "."} } {}

Discussion for (b)

The negative angle means that the velocity is 50 . size 12{"50" "." 1°} {} below the horizontal. This result is consistent with the fact that the final vertical velocity is negative and hence downward—as you would expect because the final altitude is 20.0 m lower than the initial altitude. (See [link] .)

One of the most important things illustrated by projectile motion is that vertical and horizontal motions are independent of each other. Galileo was the first person to fully comprehend this characteristic. He used it to predict the range of a projectile. On level ground, we define range    to be the horizontal distance R size 12{R} {} traveled by a projectile. Galileo and many others were interested in the range of projectiles primarily for military purposes—such as aiming cannons. However, investigating the range of projectiles can shed light on other interesting phenomena, such as the orbits of satellites around the Earth. Let us consider projectile range further.

Part a of the figure shows three different trajectories of projectiles on level ground. In each case the projectiles makes an angle of forty five degrees with the horizontal axis. The first projectile of initial velocity thirty meters per second travels a horizontal distance of R equal to ninety one point eight meters. The second projectile of initial velocity forty meters per second travels a horizontal distance of R equal to one hundred sixty three meters. The third projectile of initial velocity fifty meters per second travels a horizontal distance of R equal to two hundred fifty five meters.
Trajectories of projectiles on level ground. (a) The greater the initial speed v 0 size 12{v rSub { size 8{0} } } {} , the greater the range for a given initial angle. (b) The effect of initial angle θ 0 size 12{θ rSub { size 8{0} } } {} on the range of a projectile with a given initial speed. Note that the range is the same for 15º size 12{"15"°} {} and 75º size 12{"75°"} {} , although the maximum heights of those paths are different.

How does the initial velocity of a projectile affect its range? Obviously, the greater the initial speed v 0 size 12{v rSub { size 8{0} } } {} , the greater the range, as shown in [link] (a). The initial angle θ 0 size 12{θ rSub { size 8{0} } } {} also has a dramatic effect on the range, as illustrated in [link] (b). For a fixed initial speed, such as might be produced by a cannon, the maximum range is obtained with θ 0 = 45º size 12{θ rSub { size 8{0} }  = "45º"} {} . This is true only for conditions neglecting air resistance. If air resistance is considered, the maximum angle is approximately 38º size 12{"38º"} {} . Interestingly, for every initial angle except 45º size 12{"45º"} {} , there are two angles that give the same range—the sum of those angles is 90º size 12{"90º"} {} . The range also depends on the value of the acceleration of gravity g size 12{g} {} . The lunar astronaut Alan Shepherd was able to drive a golf ball a great distance on the Moon because gravity is weaker there. The range R size 12{R} {} of a projectile on level ground for which air resistance is negligible is given by

Questions & Answers

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
how to know photocatalytic properties of tio2 nanoparticles...what to do now
Akash Reply
it is a goid question and i want to know the answer as well
characteristics of micro business
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 ?
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.
what is the actual application of fullerenes nowadays?
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.
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.
is Bucky paper clear?
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.
Do you know which machine is used to that process?
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
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, Cc test coll. OpenStax CNX. Dec 15, 2015 Download for free at http://legacy.cnx.org/content/col11717/1.4
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