SOLUTION: 2. Suppose that a ball is thrown upward with an initial velocity of 22 ft/sec. If the ball is released at a height of 5 ft., the height equation may be written as h=-16t^2+22t+5 Wh

Question 93461: 2. Suppose that a ball is thrown upward with an initial velocity of 22 ft/sec. If the ball is released at a height of 5 ft., the height equation may be written as h=-16t^2+22t+5 When, to the nearest thousandth of a sec., will the ball be at a height of 9ft. on the way up?
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

To find the time the ball is in the air at 9 ft, simply let h=9

Subtract 9 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=22, and c=-4

Square 22 to get 484

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 possible solutions are:
or

But since we only want the height when the ball is going up, we only care about the first solution.

So our solution is which is 0.216 to the nearest thousandth
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