The maximum height reached by an object thrown directly upward is dire

Question

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.

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 .

ManojReddy · Accepted Answer

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(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

abhinavgaur · Answer

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 :?:

reto · Answer

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

lipsi18 · Answer

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

Thanks,

karant · Answer

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

vishwaprakash · Answer

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

Bunuel · Answer

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.

OptimusPrepJanielle · Answer

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

PratikHP · Answer

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,  h max = 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

mathivanan · Answer

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

mvictor · Answer

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.

sahilsnpt · Answer

Isn't the question ambiguous? Because it derives a corelation between height and velocity and asks us about the speed. We all know that speed and velocity are totally different terms.Am I correct?  Bunuel  please help ..

abhimahna · Answer

Hey sahilsnpt , Lol, I could see you are a Physics student.

Yeah, it should be velocity. I have modified the original question. Thank you!

Paras96 · Answer

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.

Let H = maximum height reached by an object thrown directly upward,
v= the velocity with which the object is thrown
k= constant

therefore, H = Kv^2
for 2 difference scenarious

=> H1/H2 = v1^2/v2^2
=> 4/9 = 16^2/v2^2
=> V2^2 = 64*9 
=> V2 = 24

Hence D

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The maximum height reached by an object thrown directly upward is dire

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