An object thrown directly upward is at a height of h feet after t seconds; where h = -16(t-3)^2 + 150. At what height, in feet, is the object 2 seconds after it reaches its maximum height?
a) 6
b) 86
c) 134
d) 150
e) 166
Many thanks
Vitto
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- vittovangind
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The formula h = -16 (t - 3)² + 150 allows us to determine the height of the object at any time. For what value of t is -16(t-3)² + 150 maximized (in other words, the object is at its maximum height)?vittovangind wrote:An object thrown directly upward is at a height of h feet after t seconds; where h = -16(t-3)² + 150. At what height, in feet, is the object 2 seconds after it reaches its maximum height?
a) 6
b) 86
c) 134
d) 150
e) 166
It might be easier to answer this question if we rewrite the formula as h = 150 - 16(t-3)²
To MAXIMIZE the value of h, we need to MINIMIZE the value of 16(t-3)² and this means minimizing the value of (t-3)²
As you can see,(t-3)² is minimized when t = 3.
We want to know the height 2 seconds AFTER the object's height is maximized, so we want to know that height at 5 seconds (3+2)
At t = 5, the height = 150 - 16(5 - 3)²
= 150 - 16(2)²
= 150 - 64
= 86
Answer: B
Cheers,
Brent
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H = 150 - 16(t-3)^2vittovangind wrote:An object thrown directly upward is at a height of h feet after t seconds; where h = -16(t-3)^2 + 150. At what height, in feet, is the object 2 seconds after it reaches its maximum height?
a) 6
b) 86
c) 134
d) 150
e) 166
Many thanks
Vitto
Try to minimize z to maximize H
T will be minimum at t = 3
But the question asks - At what height, in feet, is the object 2 seconds after it reaches its maximum height?
So after 2 seconds of maximum height the height of the object the time T will be 5
So the height of the object after 2 seconds after reaching maximum height is -
H = 150 - 16(5-3)^2
H = 150 - 16*4
H = 150 - 64
H = 86
Hence Answer must be (B)...
Abhishek
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