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Manager  Joined: 12 Oct 2009
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An object thrown directly upward is at a height of h feet  [#permalink]

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Question Stats: 68% (01:46) correct 32% (01:55) wrong based on 2591 sessions

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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. 214
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Joined: 02 Sep 2009
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Re: An object thrown directly upward is at a height of h feet  [#permalink]

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IEsailor 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. 214

Given that $$h=150-16(t-3)^2$$.

In order to maximize $$h$$, we need to minimize $$16(t-3)^2$$ (since it's subtracted from 150), which means minimizing $$(t-3)^2$$. Now, since $$(t-3)^2$$ is always non-negative, then the smallest possible value of $$(t-3)^2$$ is 0, for $$t=3$$.

Two seconds later or for $$t=3+2=5$$, the height will be $$h=150-16(5-3)^2=86$$.

Hope it's clear.
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Re: PS - Height of object  [#permalink]

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7
2
Maximum height is reached at t=3 i.e 150

2 sec's after reaching maximum height object will be in free fall mode.
=> height at t =1, is the height of the object 2 seconds after it reaches its maximum height?

substituting t=2 we have 150-16(4)
=86

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Re: PS - Height of object  [#permalink]

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5
2
max height at t=3, 150

2 sec after t=3 means t=5.

substituting h= -64 + 150 = 86
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Re: PS - Height of object  [#permalink]

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Spidy001 wrote:
Maximum height is reached at t=3 i.e 150

2 sec's after reaching maximum height object will be in free fall mode.
=> height at t =1, is the height of the object 2 seconds after it reaches its maximum height?

substituting t=2 we have 150-16(4)
=86

How you can say the max height is reached when t=3
Intern  Joined: 12 Jan 2012
Posts: 21
GMAT 1: 720 Q49 V39 Re: PS - Height of object  [#permalink]

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divyakatas wrote:
Spidy001 wrote:
Maximum height is reached at t=3 i.e 150

2 sec's after reaching maximum height object will be in free fall mode.
=> height at t =1, is the height of the object 2 seconds after it reaches its maximum height?

substituting t=2 we have 150-16(4)
=86

How you can say the max height is reached when t=3

Method 1: -16(t-3)^2 would reduce the value of 150 and h would be maximum when -16(t-3) ^2 = 0 ie t = 3
Method 2: using derivatives dh/dt = 0.
Manager  Joined: 21 Aug 2012
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Re: An object thrown directly upward is at a height of h feet  [#permalink]

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3
1
IEsailor 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. 214

Hi,

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

Now, in the above expression whenever t is of any value except 3, the overall $$(t-3)^2$$ is always +ve.
and this +ve value is multiplies by -16.. Hence, this overall value is negative.

In order to make sure that max. height is reached, $$-16(t-3)^2$$ has to be positive, but it is never positive. SO a zero value will work to get the max. height.

So t has to be 3 to get $$-16 (t -3)^2$$ as zero, and obtain the maximum height.

Now, when t=3
h=150

The question asks for the height when time is 2 second after max height. After max. height, the object will fall.
So, the time will be 5 second. ( 3+2)

$$h = 150 -16(5-3)^2$$
h = 150 - 16*4
h = 150 -64
h = 86

Hence, the height will be 86 feet.

Thanks,
Jai

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An object thrown directly upward is at a height of h feet  [#permalink]

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1
First find out what height would be considered as 'maximum' given the equation:

Since 150 is a constant, find out what the product's maximum value can be. It is a product of a negative number and a squared parenthesis; this product would ALWAYS be negative UNLESS the parenthesis yields zero - this would be the point at which the product would have maximum value.

For t = 3, the expression would yield 0 + 150. Hence, the maximum possible height given by this equation would be 150 feet. Therefore, after reaching this height, the object would start falling. The 'start' time of falling would be after 3 seconds, so adding 2 seconds would be 5:

h = -16 x (5-3)^2 + 150
h = -16 x 4 + 150
h = 86
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: An object thrown directly upward is at a height of h feet  [#permalink]

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Hi All,

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

-16(5-3)^2 + 150 = -64 + 150 = 86; Answer B

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Re: An object thrown directly upward is at a height of h feet  [#permalink]

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1
Attached is a visual that should help.
Attachments Screen Shot 2016-05-25 at 9.42.24 PM.png [ 122.1 KiB | Viewed 36272 times ]

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Re: An object thrown directly upward is at a height of h feet  [#permalink]

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

This further simplifies to
h=-16t^2+96t+6

Now this is an downward parabola,since a is negative.

The vertex or the maximum height is given by

t=-b/2a or t=-96/2*-16 or t=3 seconds
So at t=3 sec the max height is reached.

2 seconds later what is the height?

So,t=5 sec
Substitute in the height equation to get h=86 ft. Target Test Prep Representative G
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Re: An object thrown directly upward is at a height of h feet  [#permalink]

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IEsailor 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. 214

We are given an equation h = –16(t – 3)^2 + 150, with the following information:

h = height of h 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 first focus on “-16(t - 3)^2”. We ascertain that (t – 3)^2 is always either positive or 0. However, when (t – 3)^2 is multiplied by –16, a negative number, the product will be negative. Thus, the best we can do is to have the expression -16(t - 3)^2 equal 0, which would yield the maximum value of –16(t – 3)^2 + 150. We can obtain this value by letting t = 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

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Re: An object thrown directly upward is at a height of h feet  [#permalink]

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ok here is what i did in the first attempt.....
max height reached is 150 . 2 seconds after that h = -16(2-3)^2 + 150 = -16+150 = 134. 150-134 = 16. Where did i go wrong?
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: An object thrown directly upward is at a height of h feet  [#permalink]

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The equation that the prompt gives us to work with tells us the height of the ball after T seconds have elapsed.

So, after 3 seconds (meaning T=3), the height of the ball is 150 feet.

Two seconds AFTER that would be the 5 second 'mark' (meaning T=5). In your calculation, you plugged in T=2.

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Re: An object thrown directly upward is at a height of h feet  [#permalink]

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2 seconds after reaching the maximum height, time to reach the maximum height is 3 seconds, so after 2 seconds the fall is calculated is 134. I think i understand the flaw in my calculation the formula is from the starting point only hence it cannot be applied after reaching the maximum height.
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Re: An object thrown directly upward is at a height of h feet  [#permalink]

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IEsailor 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. 214

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)?

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

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Re: An object thrown directly upward is at a height of h feet  [#permalink]

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Is this concept frequently tested in GMAT? I am finding it little difficult to understand.
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: An object thrown directly upward is at a height of h feet  [#permalink]

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Hi santro789,

This is an example of a 'limit' question - and this particular 'design' is relatively rare (you might not see it at all on Test Day). In the broader sense though, the concept isn't that difficult. You're told that there's a MAXIMUM height, so you have to think about what has to happen to achieve the highest possible height given that equation (and when you think of everything in those terms, you're really just dealing with a Number Property and using critical thinking skills - NOT 'math' skills). Even if you don't spot the pattern here, you can still 'brute force' the solution: just plug in increasing values of T until you get the answer.

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An object thrown directly upward is at a height of h feet  [#permalink]

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I prepared a youtube video to explain this question. Hope you'll like it.

Check the explanation on this video
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Re: An object thrown directly upward is at a height of h feet  [#permalink]

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IEsailor 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. 214

Object reaches maximum height at t=3

Height at t=3+2=5 ; -16(5-3)^2+150=-64+150=86

IMO B

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