SOLUTION: Please answer this question. An object is launched upward at 45ft/Sec from a platform that is 40 feet high. What is the objects maximum height if the equation of the height(h) in

Question 258057: Please answer this question.
An object is launched upward at 45ft/Sec from a platform that is 40 feet high. What is the objects maximum height if the equation of the height(h) in terms of time (t) of the object is given by h(t) = -16t^2 + 45t + 40?
Found 3 solutions by stanbon, Greenfinch, drk:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
An object is launched upward at 45ft/Sec from a platform that is 40 feet high. What is the object's maximum height if the equation of the height(h) in terms of time (t) of the object is given by h(t) = -16t^2 + 45t + 40?
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max occurs when x = -b/2a = -45/(2*-16) = 1.40625
The max height is f(1.40625) = -16(-1.40625)^2 + 45(1.40625) + 40
= 71.64 ft.
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Cheers,
Stan H.
Answer by Greenfinch(383)   (Show Source): You can put this solution on YOUR website!
h = 16t^2 + 45t + 40
dh/dt = 32t + 45. When dh/dt = 0, h is a stationary point
this is at t = 45/32 = 1.406 seconds
so h = 32 x(1.406^2 + 45x 1.406 + 40 = 63.25 + 63.27 + 40 = 166.52 feet
Answer by drk(1908)   (Show Source): You can put this solution on YOUR website!
We can find the maximum height by completing the square.
step 1 - rewrite the equation as

The blanks are waiting for numbers.
step 2 - factor out a -16 only from the first two parts to get

step 3 - take 1/2 of the middle term and square it as

step 4 - multiply this squared answer by -16 to get 31.64. place this number in the second blank to get

step 5 - rewrite the trinomial as a binomial squared and simplify as

Now, the max ht occurs at 45/32 seconds @ 71.64 ft.
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