SOLUTION: 16) a ball projected vertically upward from the top of a tower has height (in feet) h(t) = -16t^2 + 64t + 80 after t seconds. At what time will the ball hit the ground?

Question 122530: 16) a ball projected vertically upward from the top of a tower has height (in feet)
h(t) = -16t^2 + 64t + 80 after t seconds. At what time will the ball hit the ground?

a) t=2
b) t=5
c) t=4
d) t=1
e) none of these

Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
When the ball hits the ground, this means (in other words, the height is zero)

Start with the given equation

Plug in . In other words, replace h(t) with 0

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=64, and c=80

Square 64 to get 4096

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

Lets look at the first part:

Add the terms in the numerator
Divide

So one answer is

Now lets look at the second part:

Subtract the terms in the numerator
Divide

So another answer is

So our solutions are:
or

However, since a negative time doesn't make sense, our only solution is

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

Answer:

So at 5 seconds, the ball will hit the ground.
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