SOLUTION: The path of a falling object is given by using the function s= -16t^2+ v sub0 t+ s sub0 where v sub0 represents the initial height in ft/sec and s sub0 represents the initial h

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Question 150685: The path of a falling object is given by using the function s= -16t^2+ v sub0 t+ s sub0 where
v sub0 represents the initial height in ft/sec and
s sub0 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...
I got.. -16t^2 + 40t + 30
B.How high is the rock after 2 seconds?
I got..
s(t)= -16(2)^2 + 40(2) + 30
s(t)= 46
C. After how many seconds will the graph reach maximum height? Show work algebraically.

D What is the maximum height?


I need help on C and D... and please tell me if I have figured A and B correctly... Thank you so very much.

Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!
Parts A and B are correct.

C)


From , we can see that , , and .

In order to find out when the rock will reach the max height, use this formula: .


Start with the given formula.


Plug in and .


Multiply 2 and to get .


Reduce.

So the rock will reach the max height when or . So the rock takes 1.25 seconds to reach the peak.

--------------------------
D)

Start with the given equation.


Plug in .


Square to get .


Multiply and to get .


Multiply and to get .


Combine like terms.

So at the peak, the rock is 55 feet high. So the max height is 55 feet.

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