SOLUTION: 7) The path of a falling object is given by the function where represents the initial velocity in ft/sec and represents the initial height in feet. Also, s represents the heig

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Question 150630: 7) The path of a falling object is given by the function where represents the initial velocity in ft/sec and represents the initial height in feet. Also, s represents the height in feet of the object at any time, t, which is measured in seconds.
a) If a rock is thrown upward with an initial velocity of 40 feet per second from the top of a 30-foot building, write the height (s) equation using this information.
Typing hint: Type t-squared as t^2
Answer:
b) How high is the rock after 2 seconds?
Answer:
Show your work here:

c) After how many seconds will the graph reach maximum height? Show your work algebraically.

Answer:
Show your work here:

d) What is the maximum height?
Answer:
Show your work here:

I have tried numerous approaches but don't feel I have the right answer. If I use the #2 in place of x my solution comes out to s=46 which does not sound reasonable to me. Please help
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
The path of a falling object is given by the function where represents the initial velocity in ft/sec and represents the initial height in feet. Also, s represents the height in feet of the object at any time, t, which is measured in seconds.
a) If a rock is thrown upward with an initial velocity of 40 feet per second from the top of a 30-foot building, write the height (s) equation using this information.
Typing hint: Type t-squared as t^2
Answer:
:
The equation consists of three parts,
Gravity: -16t^2; negative because it's pulling down
Initial velocity: +40t; plus because it's thrown upward
Initial height: 30; from a 30 ft building
:
s(t) = -16t^2 + 40t + 30
;
:
b) How high is the rock after 2 seconds?
Answer:
:
Substitute 2 for t in the above equation:
h = -16(2^2) + 40(2) + 30
h = -64 + 80 + 30
h = +46 ft; so you're right!!
;
:
c) After how many seconds will the graph reach maximum height? Show your work algebraically.
:
The max height will occur at the axis of symmetry x = %28-b%29%2F%282a%29
In your equation a=-16 and b=40
:
t = %28-40%29%2F%282%2A-16%29
t = %28-40%29%2F%28-32%29
t = 1.25 seconds it will be max height
:
:
d) What is the maximum height?
:
Substitute 1.25 for t and find the height
h = -16(1.25^2) + 40(1.25) + 30
h = -16(1.5625) + 50 + 30
h = -25
h = 55 ft is the max height