SOLUTION: A baseball is thrown with a vertical velocity of 50ft/s from an initial height of 6 ft. The height in feet of the baseball can be modeled by h(t)= -16t^2 + 50t + 6, where t is the

Question 806965: A baseball is thrown with a vertical velocity of 50ft/s from an initial height of 6 ft. The height in feet of the baseball can be modeled by h(t)= -16t^2 + 50t + 6, where t is the time in seconds since the ball was thrown. Approximately how many seconds does it take the ball to reach its maximum height? What is the maximum height that the ball reaches?
-16t^2 + 50t + 6
So first I did the following:
-b/2a= -50/2(-16)=-50/-32=25/16
Then I substituted it:
-16(25/16)^2 + 50(25/16) - 16
My teacher got:
1.5626 sec
45.06 feet(height)
However when I substituted it I got:
-16(25/16)^2 + 50(25/16) - 16
-39.0625 + 78.125 -16
23.0625
What am I doing wrong?
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
The vertex is
(25/16, 721/16)=(1.5625, 45.0625)
your 25/16 is correct. It is the exact answer.
-16t^2 + 50t + 6
-16(25/16)^2 + 50(25/16) +6
You have
-16(25/16)^2 + 50(25/16) - 16
where did you get the final -16
that changes the height by 22

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