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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

find the 15th term of the geometric sequince whose first is 18 and last term of 387
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The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
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Abhi
how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
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ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
Abhi
is it a question of log
Abhi
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Abhi
Commplementary angles
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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AMJAD
what is system testing
AMJAD
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Stotaw
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
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I'm interested in Nanotube
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the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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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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