4.5.36 A smoke jumper jumps from a plane that is 1500 ft above the ground. The function h= - 16t^2+ 1500 gives the jumper's height h in feet during the free fall att seconds
a. How long is the jumper free fall if the parachute opens at 1000 ft?
b. How long is the jumper in free fall if the parachute opens at 940 ft?
c. What is a reasonable domain and range for the function h?

We have a jumper that jumps from 1500ft above the ground, such that its height as a function of time is given by:

h(t) = -16*t^2 + 1500

The solutions are:

A) 5.59 seconds

B) 5.91 seconds

C) D = (0s, 9.68s)

    R = [0ft, 1500ft]

Notice that when t = 0, h = 1500, then at the initial time (t = 0) the height is 1500 (in ft), which means that the zero of the height is the ground.

a) How long is the jumper free fall if the parachute opens at 1000 ft?

Here we need to solve:

h = 1000 = -16*t^2 + 1500

Now we just need to solve this for t:

16*t^2 = 1500 - 1000 = 500

t^2 = 500/16 = 31.25

t = √31.25 = 5.59

So it takes 5.59 seconds to open the parachute.

B) similar to before, this time we need to solve:

h = 940 = -16*t^2 + 1500

16*t^2  =1500 - 940 = 560

t^2 = 560/16 = 35

t = √35 = 5.91

So it takes 5.91 seconds to open the parachute.

C) The domain is the set of inputs that we can use in the function.

The minimum of the domain is t = 0, when the jumper jumps.

To find the maximum of the domain we need to find the maximum time at which he could open the parachute, that is when h = 0ft, just in the ground.

Then we solve:

h = 0 = -16*t^2 + 1500

16*t^2 = 1500

t = √(1500/16) = 9.68

Then a reasonable domain is (0s, 9.68s)

And the range is the set of possible outputs, which will be all the possible heights of the jumper. We know that all of these are between 0ft and 1500ft, thus the range is [0ft, 1500ft]

If you want to learn more, you can read:

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12/11c

or

12

c

11

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!