SOLUTION: A baseball is thrown straight upwards with an initial velocity of 224 ft/s. At what time(s) does the ball reach 832 feet?

Question 299593: A baseball is thrown straight upwards with an initial velocity of 224 ft/s.
At what time(s) does the ball reach 832 feet?

Found 2 solutions by helpnalgebra, Alan3354:
Answer by helpnalgebra(91) (Show Source): You can put this solution on YOUR website!
you just divide the 224 into 832 = 3
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
A baseball is thrown straight upwards with an initial velocity of 224 ft/s.
At what time(s) does the ball reach 832 feet?
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h(t) = -16t^2 + 224t gives the height in feet as a function of t in seconds.
832 = -16t^2 + 224t
-t^2 + 14t - 52 = 0

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

The discriminant -12 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.

In the field of imaginary numbers, the square root of -12 is + or - .

The solution is , or
Here's your graph:

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Looks like it's on May 32nd.
It doesn't go that high, it's max height is 784 feet at 7 seconds.

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