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# ​Help with algebra two questions!

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Help with algebra two questions!

Guest Oct 28, 2017
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The height in meters of a rocket after  t  seconds can be modeled by

h(t)  =  -4.9(t - 4)2 + 80

a)     You can see this as the equation of a parabola in vertex form:  y = a(x - h)2 + k

So the vertex is  (4, 80) , and it is a maximum.

The maximum height is 80 meters , and it occurs when t = 4 seconds.

Also you can see that any possible value other than  0  for  -4.9(t - 4)2   will take away from the  80 .

b)     The y-intercept occurs when t = 0 .

h(0)  =  -4.9(0 - 4)2 + 80

h(0)  =  1.6

So the y-intercept is  1.6 .

It is the height of the rocket at 0 seconds – the height when it was launched.

c)     Let's find  t  such that  h(t) = 60 .

-4.9(t - 4)2 + 80  =  60       Subtract  80  from both sides

-4.9(t - 4)2  =  -20             Divide both sides by  -4.9 .

(t - 4)2  =  200/49             Take the ± square root of both sides.

t - 4  =  ±√200 / 7              Add  4  to both sides.

t  =  ±√200 / 7  +  4

t  ≈  6.02    and    t  ≈  1.98

Any value for  t  that falls between  1.98  and  6.02  will cause  h  to be greater than  60 .

6.02 - 1.98  =  4.04  . So it will be above 60 meters for about  4.04 seconds.

d)     average rate of change  =  change in height / change in time

avg rate of change  =  [  h(5.5) - h(2.5)  ] / [ 5.5 - 2.5 ]

=  [  ( -4.9(5.5 - 4)2 + 80 ) - ( -4.9(2.5 - 4)2 + 80 )  ] / [ 5.5 - 2.5 ]

=  [  ( -4.9(1.5)2 + 80 ) - ( -4.9(-1.5)2 + 80 )  ] / [ 3 ]

Notice that the two terms in the numerator will be the same, so their difference is  0 .

=  [ 0 ] / [ 3 ]

=  0

e)   This one is done the same as the last problem.

avg rate of change  =  [ h(6) - h(3) ] / [6 - 3]

=  [  ( -4.9(6 - 4)2 + 80 ) - ( -4.9(3 - 4)2 + 80 )  ] / [ 6 - 3 ]

Can you finish it from here?

f)  The rocket hits the ground when  h(t)  =  0  . That is..when

-4.9(t - 4)2 + 80  =  0

All you need to do is solve this equation for  t . If you get  8 (or something a little bigger than 8)  as a solution, then Perry is right...if not, Perry is wrong.

hectictar  Oct 29, 2017

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