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The maximum height reached by an object thrown directly upward is dire [#permalink]
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23 Jun 2015, 23:13
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Re: The maximum height reached by an object thrown directly upward is dire [#permalink]
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24 Jun 2015, 00:37
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Is the answer D? Given that the Height is directly proportional to the square of the velocity, I created the following equation, plugged in the values and solved. H1/(V1)^2 = H2/(V2)^2



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The maximum height reached by an object thrown directly upward is dire [#permalink]
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24 Jun 2015, 01:01
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The maximum height reached by an object thrown directly upward is directly proportional to the square of the velocity with which the object is thrown. If an object thrown upward at 16 feet per second reaches a maximum height of 4 feet, with what speed must the object be thrown upward to reach a maximum height of 9 feet?
height(h) is directly proportional to square of the velocity(v). h directly proportional to \(v^2\) h=k\(v^2\) where k is a constant \(\frac{h}{v^2}\) = k from the above, we can derive below equations \(\frac{h1}{(v1)^2}\)= \(\frac{h2}{(v2)^2}\)
Substituting the values (h1=4,v1=16,h2=9) \(\frac{4}{16^2}\) = \(\frac{9}{(v2)^2}\) v2=24 feet per second
Ans:D



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Re: The maximum height reached by an object thrown directly upward is dire [#permalink]
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24 Jun 2015, 05:25
Bunuel wrote: The maximum height reached by an object thrown directly upward is directly proportional to the square of the velocity with which the object is thrown. If an object thrown upward at 16 feet per second reaches a maximum height of 4 feet, with what speed must the object be thrown upward to reach a maximum height of 9 feet?
A. 12 feet per second B. 16 feet per second C. 18 feet per second D. 24 feet per second E. 48 feet per second
Kudos for a correct solution. v^2 is directly proportional to max height 16^2 is directly proportional to 4 256 is directly proportional to 4 >>> 256 / 4 = 64 >>> Now we are looking for max height of 9 >>> 9*64 = 576 (24^2 = 576) Therefore 24^2 is directly proportional to max height of 9 Answer D
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Re: The maximum height reached by an object thrown directly upward is dire [#permalink]
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24 Jun 2015, 08:48
As per given information assume h1 = max height of object at first throw, v1=velocity of the object on first throw; h2 = max height of object at second throw, v2=velocity of the object on second throw H is proportional to velocity ^2 => h1/h2= (v1/v2)^2 => 4/9= (16/x)^2 => 2/3=16/x => x=24 Hence answer is D
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Re: The maximum height reached by an object thrown directly upward is dire [#permalink]
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24 Jun 2015, 10:50
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Height is proportional to the square of velocity
h = k * v^2 (k is the constant)
4 = k * 16 ^ 2
or, k = 4 / 16 ^2
now question asking what will be speed required to attain 9 feet
9 = 4/16^2 * v^2 (Putting the value of K)
or v^2 = 9 * 16^2 / 4
v = 3*16/2 (taking under root both sides)
v = 24 feet / sec
hence, answer is D



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Re: The maximum height reached by an object thrown directly upward is dire [#permalink]
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26 Jun 2015, 06:59
Bunuel wrote: The maximum height reached by an object thrown directly upward is directly proportional to the square of the velocity with which the object is thrown. If an object thrown upward at 16 feet per second reaches a maximum height of 4 feet, with what speed must the object be thrown upward to reach a maximum height of 9 feet?
A. 12 feet per second B. 16 feet per second C. 18 feet per second D. 24 feet per second E. 48 feet per second
Kudos for a correct solution. Height is directly proportional to square of Velocity H1 / H2 = (V1)^2 / (V2)^2 4/9 = 16^2 / x^2 Solving we get x = 24 Option D



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Re: The maximum height reached by an object thrown directly upward is dire [#permalink]
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29 Jun 2015, 05:10
Bunuel wrote: The maximum height reached by an object thrown directly upward is directly proportional to the square of the velocity with which the object is thrown. If an object thrown upward at 16 feet per second reaches a maximum height of 4 feet, with what speed must the object be thrown upward to reach a maximum height of 9 feet?
A. 12 feet per second B. 16 feet per second C. 18 feet per second D. 24 feet per second E. 48 feet per second
Kudos for a correct solution. MANHATTAN GMAT OFFICIAL SOLUTION:Typically with direct proportion problems, you will be given “before” and “after” values. Simply set up ratios to solve the problem. For example, \(\frac{y_1}{x_1}\) can be used for the “before” values and , \(\frac{y_2}{x_2}\) can be used for the “after” values. You then write, \(\frac{y_1}{x_1}=\frac{y_2}{x_2}\), since both ratios are equal to the same constant k. Finally, you solve for the unknowns. In the problem given above, be sure to note that the direct proportion is between the height and the square of the velocity, not the velocity itself. Therefore, write the proportion as \(\frac{h_1}{(v_1)^2}=\frac{h_2}{(v_2)^2}\). Substitute the known values h1 = 4, v1 = 16, and h2 = 9: \(\frac{4}{(16)^2}=\frac{9}{(v_2)^2}\) \(v_2=24\). The object must be thrown upward at 24 feet per second. Answer: D.
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Re: The maximum height reached by an object thrown directly upward is dire [#permalink]
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30 Jun 2015, 07:28
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The maximum height reached by an object thrown directly upward is directly proportional to the square of the velocity with which the object is thrown. If an object thrown upward at 16 feet per second reaches a maximum height of 4 feet, with what speed must the object be thrown upward to reach a maximum height of 9 feet? h = height, x = constant of proportionality, v = velocity First solve for the constant. xh=v^2. x(4)=16^2 => x = 64 Second solve for v. 64(9)=v^2 A. 12 feet per second B. 16 feet per second C. 18 feet per second D. 24 feet per second E. 48 feet per second



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The maximum height reached by an object thrown directly upward is dire [#permalink]
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14 Aug 2015, 10:34
Unitary method for directly/indirectly proportional problems: Theory:Let the max. height reached by the object be \(h\). Let the velocity with which the object is thrown be \(v\). Given, \(h>v^2\). Now, hmax= 4 when the object is thrown at the speed (velocity of the object) of \(16 ft/sec\). For \(h = 9\) the \(speed\) \(v^2 = ?\) \(\frac{16*16*9}{4} = v^2\) Ans: 24  D
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Re: The maximum height reached by an object thrown directly upward is dire [#permalink]
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14 Aug 2015, 22:52
Bunuel wrote: The maximum height reached by an object thrown directly upward is directly proportional to the square of the velocity with which the object is thrown. If an object thrown upward at 16 feet per second reaches a maximum height of 4 feet, with what speed must the object be thrown upward to reach a maximum height of 9 feet?
A. 12 feet per second B. 16 feet per second C. 18 feet per second D. 24 feet per second E. 48 feet per second
Kudos for a correct solution. h =k*(v^2) 4=k*(16^2) k=1/64 9=(v^2)*(1/64) v^2 = 9*64 v=3*8=24 Hence, the correct option is D



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The maximum height reached by an object thrown directly upward is dire [#permalink]
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13 Nov 2015, 18:22
i got to the answer choice by approximations 4 ft upwards = k * 16^2 or 256k = 4, out of which we can see that k is 1/64 now we have another equation: 9 ft = k * x (velocity) or 9 = x/64 x = 9*64 that is ~600. Which value of an integer is closer to this number? well, 25 squared is 625. It must be smth 20+ but less than 25 24 is thus the answer.



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