#1**0 **

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