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An object thrown directly upward is at a height of h feet
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18 May 2011, 16:42
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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?
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31 Oct 2012, 04:21
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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 \(h=150-16(5-3)^2=86\).
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?
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20 Aug 2013, 01:30
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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
KUDOS if it helped..!!!
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Re: An object thrown directly upward is at a height of h feet
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25 May 2016, 22:22
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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 [ 122.1 KiB | Viewed 36823 times ]
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Re: An object thrown directly upward is at a height of h feet
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15 Jun 2016, 12:16
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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.
Re: An object thrown directly upward is at a height of h feet
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13 Oct 2017, 11:45
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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.
An object thrown directly upward is at a height of h feet
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07 Jan 2015, 13:01
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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:
Re: An object thrown directly upward is at a height of h feet
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28 Aug 2017, 13:37
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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 Answer:
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.
How you can say the max height is reached when t=3
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.
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.
Re: An object thrown directly upward is at a height of h feet
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31 Aug 2016, 20:25
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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31 Aug 2016, 22:29
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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.
Re: An object thrown directly upward is at a height of h feet
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15 Sep 2019, 18:52
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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gmatclubot
Re: An object thrown directly upward is at a height of h feet
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15 Sep 2019, 18:52