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An object was thrown upward from the top of a building. The object tra [#permalink]
 
sgpk242 wrote:
gmatophobia can you please explain how your statement 2 answer addresses the variable b?

­
sgpk242 Thanks for your question

We know that b is a positive value. For a constant value of maximum height reached and a fixed value of 'b' the time taken to reach that height would be same. We don't have to bother about the time taken here. In my explanation, I have assumed that the maximum height reached by the object is 40 meters (the calculation of maximum height includes the value of b, as h(t) depends on b). To explain further let's take the below cases

Case 1:

Assume that \(b = 10, c = 39\). The maximum height that the object reaches is 40 m.

\(h(t) = -4.9t^2 + 10t +38 \)

\(40 = -4.9t^2 + 10t + 38\)

Let's assume that the time taken to reach the maximum height = \(t_\text{max}\). We don't need to bother about the time taken to reach the maximum height as that's not relevant to the question. The time taken would be some positive value. 

As \(c < b\), at some point in time, say \(t_1\), in which \(t_1 < t_\text{max}\) the object would be at a height of 39 meters above ground. 

The object again attains a height of 39 meters while moving towards the ground. Let's assume the time now is \(t_2\). In this case \(t_2 > t_\text{max}\)

In this case, the answer to the question "Was the height of the object above the ground equal to c meters at most once?"­ is No. This is because the object attains a height of c meters (c is 39 in our case) twice. Once while going up and the second time while moving towards the ground.

Case 2:

Assume that \(b = 10, c = 31\). The maximum height that the object reaches is 40 m.

\(h(t) = -4.9t^2 + 10t +38 \)

\(40 = -4.9t^2 + 10t + 38\)

Similar to Case 1, the time taken to reach the maximum height = \(t_\text{max}\)

The object reaches the height of c (c = 31) meters only once, i.e. when moving toward the ground. 

In this case, the answer to the question " Was the height of the object above the ground equal to c meters at most once?"­ is Yes. This is because the object attains a height of c meters (c = 39 in our case) only once.

Visualization of both cases:

Attachment:
Screenshot 2024-03-26 105040.png
Screenshot 2024-03-26 105040.png [ 76.95 KiB | Viewed 936 times ]


Summary:  The value of \(b\) doesn't matter as for a fixed value of \(b\) the time taken to reach the maximum height will be the same. This problem can be easily solved by visualization and not getting into too many details of mathematics.

Hope this helped (or did I confuse you more   :lol: ). Feel free to let me know if you have further questions. 

 ­
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Re: An object was thrown upward from the top of a building. The object tra [#permalink]
gmatophobia wrote:
sgpk242 wrote:
gmatophobia can you please explain how your statement 2 answer addresses the variable b?

­
sgpk242 Thanks for your question

We know that b is a positive value. For a constant value of maximum height reached and a fixed value of 'b' the time taken to reach that height would be same. We don't have to bother about the time taken here. In my explanation, I have assumed that the maximum height reached by the object is 40 meters (the calculation of maximum height includes the value of b, as h(t) depends on b). To explain further let's take the below cases

Case 1:

Assume that \(b = 100, c = 39\). The maximum height that the object reaches is 40 m.

\(h(t) = -4.9t^2 + 100t +38 \)

\(40 = -4.9t^2 + 100t + 38\)

Let's assume that the time taken to reach the maximum height = \(t_\text{max}\). We don't need to bother about the time taken to reach the maximum height as that's not relevant to the question. The time taken would be some positive value. 

As \(c < b\), at some point in time, say \(t_1\), in which \(t_1 < t_\text{max}\) the object would be at a height of 39 meters above ground. 

The object again attains a height of 39 meters while moving towards the ground. Let's assume the time now is \(t_2\). In this case \(t_2 > t_\text{max}\)

In this case, the answer to the question "Was the height of the object above the ground equal to c meters at most once?"­ is No. This is because the object attains a height of c meters (c is 39 in our case) twice. Once while going up and the second time while moving towards the ground.

Case 2:

Assume that \(b = 100, c = 31\). The maximum height that the object reaches is 40 m.

\(h(t) = -4.9t^2 + 100t +38 \)

\(40 = -4.9t^2 + 100t + 38\)

Similar to Case 1, the time taken to reach the maximum height = \(t_\text{max}\)

The object reaches the height of c (c = 31) meters only once, i.e. when moving toward the ground. 

In this case, the answer to the question " Was the height of the object above the ground equal to c meters at most once?"­ is Yes. This is because the object attains a height of c meters (c = 39 in our case) only once.

Visualization of both cases:

Attachment:
Screenshot 2024-03-26 105040.png


Summary:  The value of \(b\) doesn't matter as for a fixed value of \(b\) the time taken to reach the maximum height will be the same. This problem can be easily solved by visualization and not getting into too many details of mathematics.

Hope this helped (or did I confuse you more   :lol: ). Feel free to let me know if you have further questions. 

 ­

­Hi gmatophobia
Why have you considered ''b > c'' in your explanation? Question says b < c 
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Re: An object was thrown upward from the top of a building. The object tra [#permalink]
KarishmaB Bunuel chetan2u

This one troubled me a bit, but finally got to the answer. Below was my approach. Is it okay?

h(t)=−4.9t^2+bt+38

at t = 0, h(t) = 38.
So, we can conclude that, initial height was 38

Statement:1 c < 38

Object will be thrown upward and then it will go downwards. Hence, object must be crossing a height less than 38 ONLY once.

Statement:2 b < c

at (t = b/9.8) object will be reaching its maximum height.

Hence, h(t)max = 38 + (b^2/19.6)

If b = 100, h(t) = 548

Now if c = 600, then answer to was the height of the object above the ground equal to C meters at most once? is YES
If C = 150, then answer is NO.­
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An object was thrown upward from the top of a building. The object tra [#permalink]
 
[quote="ashutosh_73"][/quote]
­Hey ashutosh_73

Thanks for your observation. I apologise for the oversight. 

Actually, in my explanation, the value of 'b' doesn't play a great role. I solved the question using visualization and the value of the constant only helps us calculate the height of the ball at any given point in time (which we are not interested in anyway :)). I believe that the question can be solved very easily if we realize the following points: 

1) The initial height, at t = 0, of the ball is 38 meters. 
2) Once the ball is thrown up, it reaches its maximum height and falls to the ground. 
The value of 'b' helps us with the height of the ball at a certain point in time, however, if we assume a maximum height, 'b' has not much role to play.

In statement 2, all we know is b < c, we don't know whether c is less than 38 or more than 38. If the value of c is less than 38, that height can be reached only once. If the value of c is > 38, the ball could reach that height more than once. 

Hope this helps. ­
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Re: An object was thrown upward from the top of a building. The object tra [#permalink]
Expert Reply
 
yrozenblum wrote:
An object was thrown upward from the top of a building. The object traveled upward until it reached its maximum height and then fell until it hit the ground next to the building. Between the time the object was thrown and when the object hit the ground, its height above level ground was modeled by the equation \(h(t) = -4.9t^2 + bt + 38\), where \(h(t)\) is the height, in meters, \(t\) is the number of seconds after the object was thrown, and \(b\) is a positive constant. During this time, was the height of the object above the ground equal to \(c\) meters at most once?

(1) \(c < 38\)

(2) \(b < c\)


Attachment:
2024-01-24_14-38-34.png

­An object is thrown from a height and position at any time t, is given by \(h(t) = -4.9t^2 + bt + 38\).

(1) \(c < 38\)
\(h(t) = -4.9t^2 + bt + 38\)
 \(h(t) = t(b-4.9t) + 38\). Thus at t=0 h(0) = 0+38 = 38.
MAx height is 38, and if c<38, the ball would have been at that height at some point of time.
Sufficient

(2) \(b < c\)
That leaves a lot for imagination.
c could easily be infinity or somewhere in between. 
Whatever be the value of b, the max height is 38.
If 38<b<c, then no.
If b<c<38, then yes.
Insufficient

A

ashutosh_73 you are correct in your approach.­
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Re: An object was thrown upward from the top of a building. The object tra [#permalink]
yrozenblum wrote:
An object was thrown upward from the top of a building. The object traveled upward until it reached its maximum height and then fell until it hit the ground next to the building. Between the time the object was thrown and when the object hit the ground, its height above level ground was modeled by the equation \(h(t) = -4.9t^2 + bt + 38\), where \(h(t)\) is the height, in meters, \(t\) is the number of seconds after the object was thrown, and \(b\) is a positive constant. During this time, was the height of the object above the ground equal to \(c\) meters at most once?

(1) \(c < 38\)

(2) \(b < c\)


Attachment:
2024-01-24_14-38-34.png

­
Got this qs in a practice test. Had skipped it. Jotting down my approach here-

Firstly the qs- Jotted a diagram like gmatophobia
Then realised that when t=0, i.e at the very beginning- the height is 38 units. So the height will increase till some point and then decrease as the object falls down.
The qs is asking me- in the entire experiment, was the object at a particular height more than once. If we look at the figure, we can realise that it can happen in the region from the zero point to the max point (once while going up and once while going down). For a height less than the initial point, the object will have that height only once, i.e when it is falling down.

Now evaluating the options
1. c<38
Voila. The statement is referring to the height when the object is falling down below the initial point. Hence the object would have been at that (c) height only once.
Sufficient

2. Here I applied some fuzzy logic. chetan2u is this fine?
I thought, look at the equation and the logic you have derived. In no way does the value of b or the relationship between b and c affect my analysis. 
Hence, insufficient.

Fin.
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An object was thrown upward from the top of a building. The object tra [#permalink]
yrozenblum wrote:
An object was thrown upward from the top of a building. The object traveled upward until it reached its maximum height and then fell until it hit the ground next to the building. Between the time the object was thrown and when the object hit the ground, its height above level ground was modeled by the equation \(h(t) = -4.9t^2 + bt + 38\), where \(h(t)\) is the height, in meters, \(t\) is the number of seconds after the object was thrown, and \(b\) is a positive constant. During this time, was the height of the object above the ground equal to \(c\) meters at most once?

(1) \(c < 38\)

(2) \(b < c\)


Attachment:
2024-01-24_14-38-34.png

­Statement 1 is sufficient because of the logic others stated.

For statement 2, (2) \(b < c\). Another quick algebraic approach: by analyzing the equation: \(h(t) = -4.9t^2 + bt + 38\), where \(h(t)\), we see that the more "b" is the taller the curve would be, why? because the coefficient of "b" is "t" which is time, hence always positive, so the higher the value of "b", the higher the relative value of \(h(t)\) for the same time "t". But, we know that to answer the question we need to ensure either c<38 or c >= maximum height. Since there is no upper limit for "b" and the lower limit is b>0, we just cannot say anything about "c" being less than 38 or c>=maximum height.

 ­
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An object was thrown upward from the top of a building. The object tra [#permalink]
@yrozenblumGiven: An object was thrown upward from the top of a building. The object traveled upward until it reached its maximum height and then fell until it hit the ground next to the building. Between the time the object was thrown and when the object hit the ground, its height above level ground was modeled by the equation \(h(t) = -4.9t^2 + bt + 38\), where \(h(t)\) is the height, in meters, \(t\) is the number of seconds after the object was thrown, and \(b\) is a positive constant.

Asked: During this time, was the height of the object above the ground equal to \(c\) meters at most once?

 \(h(t) = -4.9t^2 + bt + 38 \)

At t=0
h(t) = 38;

 \(h(t) = -4.9t^2 + bt + 38 = c\)
\(4.9t^2 - bt + (c-38) = 0\)

(1) \(c < 38\)
Since the object was thrown upwards from initial height 38 and then fell downwards, height c <38 was achieved only once while falling down.
SUFFICIENT

(2) \(b < c\)
0<b<c; Both b & c are positive and c>b.
If Discriminant b^2 - 19.6(c-38) >= 0; Feasible
But if Discriminant b^2 - 19.6(c-38) < 0; Not Feasible
If c=39; b=1; Discriminant b^2 - 19.6 < 0; Not feasible
But if c=37; Discriminant b^2 +19.6 > 0: feasible regardless of value of b
NOT SUFFICIENT

IMO A­
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