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Equations

by vinay1983 » Sun Sep 01, 2013 9:46 pm

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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) 166
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by sanjoy18 » Sun Sep 01, 2013 10:03 pm
its B for me
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by sanjoy18 » Sun Sep 01, 2013 10:05 pm
its B for me
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by sanjoy18 » Sun Sep 01, 2013 10:06 pm
its B for me
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by [email protected] » Sun Sep 01, 2013 10:47 pm
Hi vinay1983,

This is an example of a "limit" question. The question requires you to figure out the "maximum" height using a given equation.

You'll notice that the first part of the equation is -16(other things), so you'll subtract something from 150 except when that first part = 0

If t = 3 seconds, then you have -16(0)^2 + 150 = 150 feet

So, 150 feet is the maximum height.

We're asked for the height 2 seconds AFTER the maximum height, so plug in t = 5

You'll have the correct answer: [spoiler]-16(5-3)^2 + 150 = -64 + 150 = 86[/spoiler]

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by Java_85 » Mon Sep 02, 2013 8:30 am
vinay1983 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
Equations with the power of two are U shape curves! The peak of this U shape curves is the only point that the Derivative is Zero i.e. d/dt (-16(t- 3)^2 + 150)==0 ==> d/dt(t^2-6t+9)==0 ==> 2t-6=0 ==> t=3 is the time that We're at the peak of this U shape Curve, 2 seconds after peak==> t=5 ==> h(5)=86 ==> B

Since it's a U shape curve at time t=1 also the height is 86. it does not matter 2 seconds before getting to peak or after! for both the height is always the same for quadratic equations.

Hope this helps.
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by Brent@GMATPrepNow » Mon Sep 02, 2013 9:30 am
Java_85 wrote:
vinay1983 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
Equations with the power of two are U shape curves! The peak of this U shape curves is the only point that the Derivative is Zero i.e. d/dt (-16(t- 3)^2 + 150)==0 ==> d/dt(t^2-6t+9)==0 ==> 2t-6=0 ==> t=3 is the time that We're at the peak of this U shape Curve, 2 seconds after peak==> t=5 ==> h(5)=86 ==> B

Since it's a U shape curve at time t=1 also the height is 86. it does not matter 2 seconds before getting to peak or after! for both the height is always the same for quadratic equations.

Hope this helps.
Your solution is 100% correct, Java_85, but I want to point out to people (who may be experiencing painful flashbacks from Calculus 101 :-)) that the GMAT does not require any knowledge of Calculus. Sure, there may be the occasional question that can be solved using derivatives or integral, but the same question will also be solvable using non-Calculus approaches)

If we rewrite the formula as h = 150-16(t-3)², we can see that, in order to MAXIMIZE the value of h we must 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 seconds.

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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by Jeff@TargetTestPrep » Fri May 18, 2018 11:30 am
vinay1983 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
We are given an equation h = -16(t - 3)^2 + 150, with the following information:

h = height, in feet

t = number of seconds

We need to determine the height, in feet, 2 seconds after it reaches maximum height. So we first need to determine the value of t when the object's height h is the maximum. In other words, we need to determine the maximum value for this equation.

We see that since -16(t - 3)^2 is nonpositive, the maximum value of -16(t - 3)^2 + 150 is 150 when -16(t - 3)^2 = 0. We see that if -16(t - 3)^2 = 0, t must be 3.

We now know that the object reaches its maximum height at t = 3 (and the maximum height is 150 ft). However, we want the height of the object 2 seconds after it reaches the maximum height. Thus, we want the height at t = 5 since 3 + 2 = 5. Thus, we can plug in 5 for t and solve for h.

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

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

h = -16(2)^2 + 150

h = -16 x 4 + 150

h = -64 + 150

h = 86

Answer B

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by ca7ch22 » Sun Mar 10, 2019 1:58 pm
[1] Find value that minimizes -16(t-3)^2

Since (t-3)^2 will always be positive (or 0), we should minimize -16(t-3)^2 by finding a value for t that will make the expression equal 0

(t-3)^2 = 0
(t-3)(t-3) = 0
t = 3

[2] Find value of h when using t + 2

t + 2
3 + 2 = 5

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

Answer B
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