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Difficulty: 555-605 Level,   Algebra,   Min-Max Problems,                              
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Re: An object thrown directly upward is at a height of h feet [#permalink]
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Video solution from Quant Reasoning:
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Re: PS - Height of object [#permalink]
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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

Answer is B.
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Re: PS - Height of object [#permalink]
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max height at t=3, 150

2 sec after t=3 means t=5.

substituting h= -64 + 150 = 86
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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


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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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

You'll have the correct answer:
-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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Attached is a visual that should help.
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Screen Shot 2016-05-25 at 9.42.24 PM.png
Screen Shot 2016-05-25 at 9.42.24 PM.png [ 122.1 KiB | Viewed 96987 times ]

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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

Answer B
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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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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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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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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


For those who know calculus
h = -16 (t -3)^2 + 150
dh/dt = -32(t-3) = 0 at t=3
2 seconds after t=3 is t=5
h = -16(5-3)^2 + 150 = 86
Therefore, (B). Obviously, beyond the scope of GMAT
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Re: An object thrown directly upward is at a height of h feet [#permalink]
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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Re: An object thrown directly upward is at a height of h feet [#permalink]
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Hi abhibad,

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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An object thrown directly upward is at a height of h feet [#permalink]
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]
An approach suggested from another answer to find maximum points in quadratic equations

Differentiate the equation , i.e slope would be 0 at maximum height.
dy/dx = -32 (t-3)=0 => t =3

Substitute t+2 into equation to get answer
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An object thrown directly upward is at a height of h feet [#permalink]
BrentGMATPrepNow Bunuel ScottTargetTestPrep, why can't we look at the question this way?

At its maximum height, the object will be at 150 feet. The object reaches this height when it has travelled for 3 seconds. After 2 seconds, it has travelled a distance of 134 ft. So, 2 seconds after the object has reached its maximum height, the object will be 150 + 134 = 284 feet from the ground.

Why is this way of thinking not correct?
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Re: An object thrown directly upward is at a height of h feet [#permalink]
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CrushTHYGMAT wrote:
BrentGMATPrepNow Bunuel ScottTargetTestPrep, why can't we look at the question this way?

At its maximum height, the object will be at 150 feet. The object reaches this height when it has travelled for 3 seconds. After 2 seconds, it has travelled a distance of 134 ft. So, 2 seconds after the object has reached its maximum height, the object will be 150 + 134 = 284 feet from the ground.

Why is this way of thinking not correct?


Hi CrushTHYGMAT,

Based on your post, you seem to understand that the MAXIMUM height of the object is 150 feet - meaning that it cannot go any higher than that. By extension, at any point OTHER than at the 3-second mark, the height of the object must be LESS than 150 feet.

IF the question had asked for the total distance that the object had traveled (re: it went all the way "up", then started to come "down" a certain distance), then the answer would be greater than 150 feet once you pass the 3-second mark (so you would end up adding some number to 150) - but that's not what's involved in this prompt.

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