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
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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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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]
GMAT assassins aren't born, they're made,
Rich
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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 ==> Bvinay1983 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
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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Your solution is 100% correct, Java_85, but I want to point out to people (who may be experiencing painful flashbacks from Calculus 101Java_85 wrote: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 ==> Bvinay1983 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
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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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
Cheers,
Brent
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We are given an equation h = -16(t - 3)^2 + 150, with the following information: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
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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[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
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