Asked by Philip
A stone is dropped from a height of 100 feet. The time it takes for the stone to reach a height of h is given by the function t = 1/4 √(100 - h) , where t is the time in seconds. Identify the domain and range of the function, and determine the height of the stone after 2 seconds.
My work -
2 = 1/4 √(100 - h)
I multiplied both sides by 4 from -
8 = √(100 - h)
Then, I got -
8 = 10√(-h)
I divided by 10 -
8/10 = √(-h)
now what?
Do I divide by -1 on both sides?
My work -
2 = 1/4 √(100 - h)
I multiplied both sides by 4 from -
8 = √(100 - h)
Then, I got -
8 = 10√(-h)
I divided by 10 -
8/10 = √(-h)
now what?
Do I divide by -1 on both sides?
Answers
Answered by
Damon
My work -
2 = 1/4 √(100 - h)
I multiplied both sides by 4 from -
8 = √(100 - h)
------------- OK SO FAR BUT
square both sides
64 = 100 - h
h = 100-64
h = 36
domain from 100-h must be +
0<h<100
for range
if h = 0, t = 10/4 hits ground
if h = 100, t = 0 drop it then
so 0<=t<=10/4
============================
now let me check original
fell distance 100 - h = d
v = 32 t
d = 16t^2
t^2 = d/16
t = (1/4) sqrt d
t = (1/4) sqrt(100-h)
ok, valid
2 = 1/4 √(100 - h)
I multiplied both sides by 4 from -
8 = √(100 - h)
------------- OK SO FAR BUT
square both sides
64 = 100 - h
h = 100-64
h = 36
domain from 100-h must be +
0<h<100
for range
if h = 0, t = 10/4 hits ground
if h = 100, t = 0 drop it then
so 0<=t<=10/4
============================
now let me check original
fell distance 100 - h = d
v = 32 t
d = 16t^2
t^2 = d/16
t = (1/4) sqrt d
t = (1/4) sqrt(100-h)
ok, valid
Answered by
Philip
I am so dumb how did I not think of squaring.
Thanks.
Thanks.
Answered by
Damon
You are welcome.
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