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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