SOLUTION: A ball is thrown vertically upward. After t seconds, its height h (in feet) is given by the function h(t)= 56t-16tē . What is the maximum height that the ball will reach? Do n

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Question 359046: A ball is thrown vertically upward. After t seconds, its height h (in feet) is given by the function h(t)= 56t-16tē . What is the maximum height that the ball will reach?
Do not round your answer.
*here was my approach
i factored the equation getting:
= 4t(-4t+14)
= 4t = 0 -4t+14 = 0
t = 0 -4t = -14
t = 7/2
The answer i came up with was 3.5ft, is this correct?

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
No, that's not correct.
You're solving for when h=0.
h%28t%29=56t-16t%5E2
h%28t%29=8t%287-2t%29
h=0 when
t=0 and 7-2t=0
t=0 and 2t=7
t=0 and t=7%2F2
Using a symmetry argument, then the maximum height must occur when time is between these two values, t=%281%2F2%29%287%2F2%29=7%2F4
Find h%287%2F4%29
h%287%2F4%29=56%287%2F4%29-16%287%2F4%29%5E2
h%287%2F4%29=98-49
highlight%28h%287%2F4%29=49%29ft
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You could also have converted the height equation to vertex form, y=a%28t-h%29%5E2%2Bk since the maximum value occurs at the vertex.
I use y instead of h%28t%29 here because (h,k) is traditionally the vertex.
Complete the square.
y=-16t%5E2%2B56t
y=-16%28t%5E2-%287%2F2%29t%29
y=-16%28t%5E2-%287%2F2%29t%2B49%2F16%29%2B16%2A%2849%2F16%29
y=-16%28t-%287%2F4%29%29%5E2%2B49
So then the maximum value of y=49 occurs at t=7%2F4
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