SOLUTION: A man launches a projectile from top of 120' building. The height of the projectile above the ground is given by h=120+8t-16t2 (power of 2) where t = time in seconds after launch

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Question 92897: A man launches a projectile from top of 120' building. The height of the projectile above the ground is given by h=120+8t-16t2 (power of 2) where t = time in seconds after launch. WHEN WILL THE PROJECTILE BE 60 FEET OFF THE GROUND
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!


Plug in h=60

Subtract 60 from both sides

Let's use the quadratic formula to solve for t:


Starting with the general quadratic



the general solution using the quadratic equation is:



So lets solve ( notice , , and )

Plug in a=-16, b=8, and c=60



Square 8 to get 64



Multiply to get



Combine like terms in the radicand (everything under the square root)



Simplify the square root (note: If you need help with simplifying the square root, check out this solver)



Multiply 2 and -16 to get -32

So now the expression breaks down into two parts

or


Now break up the fraction


or


Simplify


or


So these expressions approximate to

or


So our solutions are:
or

Since a negative time doesn't make much sense, our only solution is



which is about 2.2 seconds

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