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An object thrown directly upward is at a height of h feet af

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An object thrown directly upward is at a height of h feet af [#permalink] New post 10 May 2020, 13:16
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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. 214


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[Reveal] Spoiler: OA

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Re: An object thrown directly upward is at a height of h feet af [#permalink] New post 10 May 2020, 13:56
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Carcass 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. 214


Kudos for the right solution and explanation


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)?

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

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Brent
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Re: An object thrown directly upward is at a height of h feet af [#permalink] New post 10 May 2020, 14:01
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Re: An object thrown directly upward is at a height of h feet af [#permalink] New post 11 May 2020, 05:54
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Given -16 (t-3)^2 + 150 and that the question is asking for maximum height, the maximum height occurs at t = 3 to zero out the - 16 (t - 3)^2

Maximum height = - 16 (3-3)^2 + 150 = 150

Now what happens after it reaches maximum height and continues for another 2 seconds, in addition to the 3 seconds to reach maximum height?

-16 (5-3)^2 + 150 = -16 (2^2) + 150 = - 16 * 4 + 150 = - 64 + 150 = 86
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Re: An object thrown directly upward is at a height of h feet af [#permalink] New post 11 May 2020, 06:43
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Equation for height of object h = 150 - 16\((t-3)^2\)

The maximum height h the object can travel will occur only when the term 16\((t-3)^2\) is zero

Hence t-3 = 0 which makes t=3

Maximum height is reached when t=3

So the time which is 2 secs after maximum height = 3 + 2 = 5

Height at 5 secs = 150 - 16\((5 - 3)^2\) = 86

Answer is B
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Re: An object thrown directly upward is at a height of h feet af [#permalink] New post 12 Oct 2020, 19:48
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The question can be rewritten as: h = 150 - 16 (t-3)^2

As (t-3)^2 always greater than 0, so -16(t-3)^2 =< 0, so 150 - 16 (t-3)^2 =< 150, so the maximum height will be 150 when (t-3)^2 = 0, which means t = 3

The question asked at the time of 2 seconds after the max height, therefore t'= 3+2 = 5. Apply to the equation, we have:

h = 150 - 16 (5-3)^2 = 150 - 16* (2^2) = 150 - 16*4 = 150-64 = 86

Therefore, the answer is B
Re: An object thrown directly upward is at a height of h feet af   [#permalink] 12 Oct 2020, 19:48
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