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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

explain and give four Example hyperbolic function
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_3_2_1
felecia
⅗ ⅔½
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_½+⅔-¾
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The denominator of a certain fraction is 9 more than the numerator. If 6 is added to both terms of the fraction, the value of the fraction becomes 2/3. Find the original fraction. 2. The sum of the least and greatest of 3 consecutive integers is 60. What are the valu
SABAL Reply
1. x + 6 2 -------------- = _ x + 9 + 6 3 x + 6 3 ----------- x -- (cross multiply) x + 15 2 3(x + 6) = 2(x + 15) 3x + 18 = 2x + 30 (-2x from both) x + 18 = 30 (-18 from both) x = 12 Test: 12 + 6 18 2 -------------- = --- = --- 12 + 9 + 6 27 3
Pawel
2. (x) + (x + 2) = 60 2x + 2 = 60 2x = 58 x = 29 29, 30, & 31
Pawel
ok
Ifeanyi
on number 2 question How did you got 2x +2
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combine like terms. x + x + 2 is same as 2x + 2
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x*x=2
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Mark and Don are planning to sell each of their marble collections at a garage sale. If Don has 1 more than 3 times the number of marbles Mark has, how many does each boy have to sell if the total number of marbles is 113?
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Mark = x,. Don = 3x + 1 x + 3x + 1 = 113 4x = 112, x = 28 Mark = 28, Don = 85, 28 + 85 = 113
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how do I set up the problem?
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hello, I am happy to help!
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please can go further on polynomials quadratic
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X=16
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Want to review on complex number 1.What are complex number 2.How to solve complex number problems.
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yes i wantt to review
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hello, if you have a question about Algebra 2. I may be able to help. I am an Algebra 2 Teacher
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4
Trista
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Sidiki Reply
can you teacch how to solve that🙏
Mark
Solve for the first variable in one of the equations, then substitute the result into the other equation. Point For: (6111,4111,−411)(6111,4111,-411) Equation Form: x=6111,y=4111,z=−411x=6111,y=4111,z=-411
Brenna
(61/11,41/11,−4/11)
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x=61/11 y=41/11 z=−4/11 x=61/11 y=41/11 z=-4/11
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Need help solving this problem (2/7)^-2
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x+2y-z=7
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what is the coefficient of -4×
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-1
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the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
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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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