Abaseball is hit with an initial upward velocity of 70 feet per second from a height of 4 feet above the ground. the equation h= −16t^2 +70t + 4 models the height in feet t seconds after it is hit. after the ball gets to its maximum height, it comes down and is caught by another player at a height of 6 feet above the ground. about how long after it was hit does it get caught?

To solve you need to set the equation equal to 6 (the height at which the player caught the ball.

6 = -16t^2 + 70t + 4

Next put the equation in standard form by subtracting 6 from both sides

-16t^2 + 70t - 2 = 0

This equation can be simplified by dividing by 2

-8t^2 + 35t - 1 = 0

This equation cannot be factored, but we can use the quadratic formula to find a value for x. Using the equation above we can find the values for a=-8, b = 35 and c = -1.

using the quadratic formula we can solve for x

-b +/- sqrt(b^2 - 4ac)

       2a

The solutions are

0.03 and 4.35. as 0.03 seems an unrealistic time to hit and catch a baseball we would expect the time to be 4.35 seconds.

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