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0.5 Quadratic equations  (Page 3/5)

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Clearly v 0 is 124 ft/s. Since the bottle rocket is launched from the surface, the value for y 0 is 0. We will substitute these values into the equation of motion for a projectile.

y ( t ) = − 1 2 g t 2 + ( 124 ) t + 0 size 12{y \( t \) = - { {1} over {2} } `g`t rSup { size 8{2} } + \( "124" \) t`+0} {}

Next, we substitute 32 ft/s2 for g. We use this value because in this exercise, we make use of the British system of units.

y ( t ) = − 1 2 ( 32 ft / s 2 ) t 2 + ( 124 ft / s ) t size 12{y \( t \) = - { {1} over {2} } ` \( "32"` ital "ft"/s rSup { size 8{2} } \) `t rSup { size 8{2} } + \( "124"` ital "ft"/s \) `t} {}

To find the height of the bottle rocket at a time 1.5 seconds after its launch, we make the substitution t = 1.5 s.

y ( 1 . 5 s ) = − 1 2 ( 32 ft / s 2 ) ( 1 . 5 s ) 2 + ( 124 ft / s ) ( 1 . 5 s ) size 12{y \( 1 "." 5`s \) = - { {1} over {2} } ` \( "32"` ital "ft"/s rSup { size 8{2} } \) ` \( 1 "." 5`s \) rSup { size 8{2} } + \( "124"` ital "ft"/s \) ` \( 1 "." 5`s \) } {}
y ( 1 . 5 s ) = − 36 ft + 186 ft = 150 ft size 12{y \( 1 "." 5`s \) = - "36"` ital "ft"+"186"` ital "ft"="150"` ital "ft"} {}

So we conclude that the bottle rocket attains a height of 150 ft at a time 1.5 seconds after launch.

Example 5: Let us enforce that same conditions on the bottle rocket as were presented in Example 4. Find the length of time it will take for the bottle rocket to strike the surface.

In order to determine the time at which the bottle rocket will strike the surface, we must find the value for t that leads to a height of 0. As before, we make use of the equation of motion

y ( t ) = − 1 2 ( 32 ft / s 2 ) t 2 + ( 124 ft / s ) t size 12{y \( t \) = - { {1} over {2} } \( "32"` ital "ft"/s rSup { size 8{2} } \) `t rSup { size 8{2} } + \( "124"` ital "ft"/s \) `t} {}

Because we are interested in determining when the bottle rocket hits the surface, we set the left hand side of the equation to 0

0 = − 1 2 ( 32 ft / s 2 ) t 2 + ( 124 ft / s ) t size 12{0= - { {1} over {2} } \( "32"` ital "ft"/s rSup { size 8{2} } \) `t rSup { size 8{2} } + \( "124"` ital "ft"/s \) `t} {}

If we restrict the units of t to seconds, we can simplify this equation

− 16 t 2 + 124 t = 0 size 12{ - "16"`t rSup { size 8{2} } +"124"`t=0} {}

The left hand side of the equation can be factored as

t × ( − 16 t + 124 ) = 0 size 12{t times \( - "16"`t+"124" \) =0} {}

The roots of this equation can be found by setting each factor to 0. That is

t = 0 size 12{t=0} {}
− 16 t + 124 = 0 size 12{ - "16"t+"124"=0} {}

As a consequence, the root associated with the first factor which is 0 corresponds to the time of the launch. The second equation yields the root, 7.75 seconds. This root corresponds to the time at which the bottle rocket strikes the surface.

Example 6: Let us consider the same bottle rocket as before, but with one major exception. In this exercise, assume that the launch site of the bottle rocket is moved to a bluff whose height is 100 ft above the surface. Find the length of time it will take for the bottle rocket to strike the surface below the bluff.

The situation is depicted in Fig 1.

Flight path of a projectile.

Once again, we will make use of the equation of motion for a projectile. However in this case we establish the term y 0 as 100 ft. This is due to the bottle rocket being launched from the bluff which is 100 ft above the surface. Thus we have

y ( t ) = − 1 2 ( 32 ft / s 2 ) t 2 + ( 124 ft / s ) t + 100 ft size 12{y \( t \) = - { {1} over {2} } \( "32"` ital "ft"/s rSup { size 8{2} } \) `t rSup { size 8{2} } + \( "124"` ital "ft"/s \) `t+"100"` ital "ft"} {}

To find the length of time in seconds at which the bottle rocket will strike the surface, we must solve for the roots of the following quadratic equation.

0 = − 16 t 2 + 124 t + 100 size 12{0= - "16"`t rSup { size 8{2} } +"124"`t+"100"} {}

The quadratic formula provides us with a simple way to determine the roots of a quadratic equation of the form

0 = a t 2 + b t + c size 12{0=a`t rSup { size 8{2} } +b`t+c} {}

The quadratic formula tells us that the roots of this equation are given in terms of the coefficients ( a , b , and c ) as follows

roots = − b ± b 2 − 4 a c 2 a size 12{ ital "roots"= { { - b` +- sqrt {b rSup { size 8{2} } - 4`a`c} } over {2`a} } } {}

For our problem we express the roots as

In the problem at hand, the values for a , b , and c are -16, 124, and 100 respectively. Substitution of these values into the quadratic formula yield

roots = − 124 ± 15 , 376 + 6, 400 − 32 size 12{ ital "roots"= { { - "124" +- sqrt {"15","376"+6,"400"} } over { - "32"} } } {}

Further arithmetic manipulation on this expression leads us to determine the two roots to be -0.736 and 8.49 seconds. The first root is not physically possible, so it can be ignored. The second root (8.49 seconds) tells us the time at which the bottle rocket will strike the surface.
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Source:  OpenStax, Math 1508 (laboratory) engineering applications of precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11337/1.3
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