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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} } \( - 1 "." "19" \) "."} {}

Thus,

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

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
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
what does nano mean?
Anassong Reply
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
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
Daniel
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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