SOLUTION: Hi! This is a off of my homework that I am having problems with. I have done the work, but am unsure if I have done it correctly. Any insight and or corrections would be greatly

Question 146025: Hi!
This is a off of my homework that I am having problems with. I have done the work, but am unsure if I have done it correctly. Any insight and or corrections would be greatly appreciated!
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.
a) If a rock is thrown upward with an initial velocity of 64 feet per second from the top of a 25-foot building, write the height (s) equation using this information.
Typing hint: Type t-squared as t^2
Answer: 16t^2+64t+25
b) How high is the rock after 1 second?
Answer: 105 ft.
Show your work here: 16t^2+64t+25
16(1)^2+64(1)+25
16(1)+64+25
16+64+25
c) After how many seconds will the graph reach maximum height?
Answer: 2
Show your work here: x=-b/2a
a=16, b=64, c=25
x=-(64)/2(16)
x=-64/32
x= -2
d) What is the maximum height?
Answer: (-2,-29)
Show your work here:
y=16t^2+64t+25
y=16(-2)^2+64(-2)+25
y=16(4)-128+25
y=74-128+25
y= -29

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Answer: 16t^2+64t+25
--------------------
That equation should be h(t) = -16t^2 + 64t + 25
That negative will change some of your conclusions.
-----------
Why is it negative.
That term expresses the downward effect of gravity on
the object while it is in the air for t seconds.
Cheers,
Stan H.
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