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555-605 (Medium)|   Algebra|      
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chandu2016
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A ball thrown up is at a height of h feet, t seconds after it was thro 18 Mar 2018, 04:53
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80% (01:53) correct 20% (02:23) wrong based on 464 sessions

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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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B
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A ball thrown up is at a height of h feet, t seconds after it was thro 18 Mar 2018, 09:30
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chandu2016
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
Show more

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A ball thrown up is at a height of h feet, t seconds after it was thro 18 Mar 2018, 05:41
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chandu2016
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
Show more

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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BrentGMATPrepNow
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A ball thrown up is at a height of h feet, t seconds after it was thro 18 Mar 2018, 07:20
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chandu2016
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
Show more

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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A ball thrown up is at a height of h feet, t seconds after it was thro 01 Sep 2025, 22:52
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