The figure above shows the position, at two instants of time, of a 10-

Question

https://gmatclub.com/forum/download/file.php?id=55205 The figure above shows the position, at two instants of time, of a 10-meter-long rod as it was falling down to the ground. If one end of the rod was pinned to the ground, by what vertical distance did the rod fall between the two instants of time A. 5(\sqrt{3} - 1) meters B. \sqrt{3} meters C. \frac{5}{\sqrt{3}} meters D. 10(\sqrt{3} - 1) meters E. 5\sqrt{3} meters 1.png

Superman249 · Answer

IMO:- A

The DIfference of Height at different Instant or vertical distance it falls will be = Length of Rod ( Sin60*-Sin30*)
 
=10 ((sqrt(3) -1)/2) => 5 (sqrt(3) -1)

manikmehta95 · Answer

https://gmatclub.com/forum/download/file.php?id=55205 
The figure above shows the position, at two instants of time, of a 10-meter-long rod as it was falling down to the ground. If one end of the rod was pinned to the ground, by what vertical distance did the rod fall between the two instants of time

A.  5(\sqrt{3} - 1)  meters

B.  \sqrt{3}  meters

C.  \frac{5}{\sqrt{3}}  meters

D.  10(\sqrt{3} - 1)  meters

E.  5\sqrt{3}   meters

Let the time at first instance be t1 and at second instance be t2. 
At t1:
vertical projection of the rod can be calculated using a 60degree angle given: initial height = 10*sin(60) = 10*(\sqrt{3}/2)
At t2:
new vertical projection is: final height = 10*sin(30) = 10*(1/2)

change in height = final - initial = 5*(\sqrt{3} - 1)

Ans - A

GMATWhizTeam · Answer

Solution 
 • Based on the given information, let us redraw the given diagram as below:
 https://www.dropbox.com/s/mn9lm60ektss1s3/vert_hei.jpg?dl=1 
 o So, we need to find the value of  AC – ED . 
• Now, triangle ABC, is a 30° -60° -90° triangle, whose hypotenuse is 10 m.
 o We know, that in a 30° – 60° - 90° triangle the ratio of corresponding sides are  1: \sqrt{3} : 2  
  So,  BC : AC : AB = 1: \sqrt{3} : 2 = 1*5 : 5*\sqrt{3} : 2*5  
 Since,  AB = 10 ⟹ AC = 5*\sqrt{3} 
 Similarly in triangle, BED, ED is the side opposite to the 30° angles with hypotenuse equal to 10 m,
 • So,  ED = 5 
• Therefore,  AC – ED = 5*\sqrt{3} – 5 = 5 (\sqrt{3} – 1)  meters    
Thus, the correct answer is  Option A.

Pawan0802 · Answer

Bunuel : the source of this question is eGMAT

Paras96 · Answer

At every angle, the hypotenuse i.e. length of rod = 10 meter and only the height from the ground will change

When the angle is 60, the perpendicular = 5√3 
When the angle is 30, the perpendicular = 5

Therefore vertical distance travelled = 5√3 - 5= 5(√3-1)

Hence A

The figure above shows the position, at two instants of time, of a 10-

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