SOLUTION: How high is the rock after 0.5 seconds? After how many seconds will the rock reach maximum height? What is the maximum height? S(t) = -16t^2 + 32t + 40 A falling object is giv

Question 90522: How high is the rock after 0.5 seconds?
After how many seconds will the rock reach maximum height?
What is the maximum height?
S(t) = -16t^2 + 32t + 40
A falling object is given by the function "S= -16t^2 + Vot + So" where "Vo" represents the initial velocity in ft/sec and "So" represents the initial height.
A baseball is thrown upward with an initial velocity of 32 feet per second from the top of a 40-foot skyscraper
32(0.5)=16ft high after 0.5 seconds
32(40ft)= 1280seconds reaches max height
40ft is max height
is this anywhere near correct?
help

Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
"How high is the rock after 0.5 seconds?"

Simply plug in t=0.5 to find the height

Lets evaluate

Start with the given polynomial

Plug in

Raise 0.5 to the second power to get 0.25

Multiply -16 by 0.25 to get -4

Multiply 32 by 0.5 to get 16

Now combine like terms

So after 0.5 seconds the rock is 52 ft high

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"After how many seconds will the rock reach maximum height?"

If we graph as a function of x we get

and with a graphing calculator (or just simply by looking at the graph) we find the peak when x=1

So after 1 second, the rock has reached the max height

--------------------------------------------------------------------

"What is the maximum height?"

Simply plug in t=1 to find the max height

Lets evaluate

Start with the given polynomial

Plug in

Raise 1 to the second power to get 1

Multiply 16 by 1 to get 16

Multiply 32 by 1 to get 32

Now combine like terms

So the max height is 56 feet
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