Hi everybody,

I am new and I hope it is ok to post this question here.

Could please someone be so kind and help in with this question.

I don't really understand the answer from the book.

Many thanks.

## OG 13th Edition Question 129, PS

##### This topic has expert replies

- neelgandham
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**Posts:**1060**Joined:**13 May 2011**Location:**Utrecht, The Netherlands**Thanked**: 318 times**Followed by:**52 members

For the benefit of other users, here is the question:

h = (

Maximum value of

h = -16 (t - 3)^2 + 150.

h is maximum when t = 3 . For any t greater than or less than 3, the value of h is less than 150.

In conclusion, when t = 3 secs, the object is at the maximum height of 150 meters.

2 seconds after it reached the maximum height, t = 5

h = -16 (5 - 3)^2 + 150 feet.

h = (-16 * 4) + 150 feet.

h = -64 + 150 feet.

h = 86 feet.

Let me know if still need help.

p.s: It is so good to be back!

h = -16 (t - 3)^2 + 150.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

h = (

**"Negative number"*****"Square of a number"**) + 150.Maximum value of

**h**is when the**"Square of a number"**is 0.h = -16 (t - 3)^2 + 150.

h is maximum when t = 3 . For any t greater than or less than 3, the value of h is less than 150.

In conclusion, when t = 3 secs, the object is at the maximum height of 150 meters.

2 seconds after it reached the maximum height, t = 5

h = -16 (5 - 3)^2 + 150 feet.

h = (-16 * 4) + 150 feet.

h = -64 + 150 feet.

h = 86 feet.

Let me know if still need help.

p.s: It is so good to be back!

Anil Gandham

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- GMATGuruNY
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Equation, rephrased: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

OA B

h = 150 - 16t - 3)Â².

To MAXIMIZE the value of h, we need to MINIMIZE the value SUBTRACTED from 150: 16(t-3)Â².

Since (t-3)Â² cannot be negative, the smallest possible value of 16(t-3)Â² is 0.

16(t-3)Â² = 0 when t=3.

Thus, the maximum height occurs when t=3.

Two seconds later, t=5.

When t=5, h = 150 - 16(5-3)Â² = 150-64 = 86.

The correct answer is B.

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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)?aman88 wrote:Q: 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

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