An object was thrown upward from the top of a building. The object tra

Question

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) \(b < c\)

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Official Answer

A

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gmatophobia · Accepted Answer

At t = 0, the height of the object = the height of the building

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

The object traveled upward until it reached its maximum height and then fell until it hit the ground next to the building

Hence, we can infer that the height of the object increased and then decreased, and finally, when it hit the ground, the height of the object was zero.

The object will attain a height more than once if that height is part of its onward and return journey. Put other words, if  c  is between 38 and the maximum height of the object , the object will attain the height ' c ' meters above ground twice. In all other cases, the object will attain the height 'c' meters above ground at most once.

Statement 1

(1)  c < 38

The height of the object first increased and then decreased. As  c  is between  0  and  38 , we can infer that at some point in time when the object was traveling downwards, the object must have attained a height equal to  c  meters. In its journey, the object attained this height only once.

The response to the question "Was the height of the object above the ground equal to  c  meters at most once?" is Yes.

The statement is sufficient. We can eliminate B, C, and E.

Statement 2

(2)  b < c

We don't know the maximum height reached by the object. Assume that the maximum height reached by the object is  40  meters and the value of  c  is 39 meters, we can conclude that the object reached the value of  c  meters above the ground twice during its journey, once when the object moved upwards and once when the object moved downwards.

In that case, the response to the question "Was the height of the object above the ground equal to  c  meters at most once?" is No.

However, if the maximum height reached by the object is  40  meters and the value of  c  equals  31  meters, we can infer that the object attainted that height only once during the downward journey.

In that case, the response to the question "Was the height of the object above the ground equal to  c  meters at most once?" is Yes.

Hence, statement 2 is not sufficient. We can eliminate D.

Option A

ashutosh_73 · Answer

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.

sgpk242 · Answer

gmatophobia  can you please explain how your statement 2 answer addresses the variable b?

gmatophobia · Answer

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

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

). Feel free to let me know if you have further questions.

). Feel free to let me know if you have further questions.

ashutosh_73 · Answer

Hi  gmatophobia 
Why have you considered  ''b > c''  in your explanation? Question says  b < c

gmatophobia · Answer

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.

chetan2u · Answer

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.

RenB · Answer

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.

8Harshitsharma · Answer

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.

Kinshook · Answer

@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

KarishmaB · Answer

h(t) = -4.9t^2 + bt + 38  This is height at any given time.

So when t = 0 (when the object is thrown), h(t) = 38. 
This means the height of the building from which the object is thrown is 38 m. The object will now gain height and then come back.

When t = 1, h(t) = b + 33 (approx)
When t = 2, h(t) = 2b + 18 (approx)
and so on. Finally h(t) will become 0 for some value of t.

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

If c < 38, then the height of the object was c at most once (while coming down)
If c is more than 38 but less than the greatest height the object reached, then that height was reached by the object twice - once while going up and then while coming down. (as shown by gmatophobia in his figures).

(1)  c < 38

Any height less than 38 was reached only once. Sufficient.

(2)  b < c   
This doesn't indicate the value of c.

Say c = 40 and b = 10
At t = 1, h(t) = 10 + 33 = 43 (height reached at t = 1)
At t = 2, h(t) = 2*10 + 18 = 38 (Now the object is coming down)
Hence the height of 40 is reached twice - once while going up and then while coming down.

But if c = 1000 and b = 10, the object never reaches the height c.

Hence was the height c reached at most once? We can't say till we don't know more about c. Not sufficient alone.

Answer (A)

luvsic · Answer

Much of the confusion behind this question seems to be from the variable "b"

b is simply a variable encoding the velocity of the ball

Intuitively, how fast you throw a ball doesn't change what actually happens (in terms of whether the ball will hit a certain height once or twice), it only changes how quickly it will happen and when it will drop to the ground, so the value of b is irrelevant

Only the value of c and its relative position to 38m matters, if it exceeds 38m, the ball would hit c twice (once when going up, and once when falling down), if c < 38m , then the ball will only hit it once when falling down

Just note that for large c, there exists values of b such that the ball may hit "c" 0 times, but this scenario wasn't explored in the question

kartickdey · Answer

if you assume b=10, then how can you assume max height is 40. if you fix the value of b, then the max height gets automatically fixed.

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