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A ball thrown up is at a height of h feet, t seconds after it was thro

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A ball thrown up is at a height of h feet, t seconds after it was thro  [#permalink]

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New post 18 Mar 2018, 04:53
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A ball thrown up is at a height of h feet, t seconds after it was thrown, where \(h=-2(t-5)^2+100\). What is the height of the ball now once it reached its maximum height and then descended for 5 seconds?

(A) 36 feet
(B) 50 feet
(C) 64 feet
(D) 72 feet
(E) 96 feet
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Re: A ball thrown up is at a height of h feet, t seconds after it was thro  [#permalink]

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New post 18 Mar 2018, 05:41
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chandu2016 wrote:
Approach please:

A ball thrown up is at a height of h feet, t seconds after it was thrown, where \(h=-2(t-5)^2+100\).
What is the height of the ball now once it reached its maximum height and then descended
for 5 seconds?
(A) 36 feet
(B) 50 feet
(C) 64 feet
(D) 72 feet
(E) 96 feet


Of course in these questions, we take the equation true for upwards or downwards..
\(h=-2(t-5)^2+100\).
Max h will be when t-5 is 0..
So h is 100 at t =5..
After 5 secs it will be\(h=-2(5+5-5)^2+100=100-2*25=50\).

B
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Re: A ball thrown up is at a height of h feet, t seconds after it was thro  [#permalink]

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New post 18 Mar 2018, 07:20
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chandu2016 wrote:
A ball thrown up is at a height of h feet, t seconds after it was thrown, where \(h=-2(t-5)^2+100\). What is the height of the ball now once it reached its maximum height and then descended for 5 seconds?

(A) 36 feet
(B) 50 feet
(C) 64 feet
(D) 72 feet
(E) 96 feet


The formula h = -2(t - 5)² + 100 allows us to determine the height of the object at any time. For what value of t is -2(t - 5)² + 100 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 = 100 - 2(t - 5)²
To MAXIMIZE the value of h, we need to MINIMIZE the value of 2(t - 5)² and this means minimizing the value of (t - 5)²
As you can see, (t - 5)² is minimized when t = 5.

We want to know the height 5 seconds AFTER the object's height is maximized, so we want to know that height at 10 seconds (5 + 5)

At t = 10, the height = 100 - 2(10 - 5)²
= 100 - 2(5)²
= 100 - 50
= 50
Answer: B

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Re: A ball thrown up is at a height of h feet, t seconds after it was thro  [#permalink]

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New post 18 Mar 2018, 09:30
1
chandu2016 wrote:
A ball thrown up is at a height of h feet, t seconds after it was thrown, where \(h=-2(t-5)^2+100\). What is the height of the ball now once it reached its maximum height and then descended for 5 seconds?

(A) 36 feet
(B) 50 feet
(C) 64 feet
(D) 72 feet
(E) 96 feet


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Re: A ball thrown up is at a height of h feet, t seconds after it was thro &nbs [#permalink] 18 Mar 2018, 09:30
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