On the equator at the at the vernal equinox, at noon a straight stick driven vertically in to the ground casts a shadow.

Question

On the equator at the at the vernal equinox, at noon a straight stick driven vertically in to the ground casts no shadow. What is the percent difference in the lengths of it's shadow measured at 1:45 PM and 2 : 30 PM. 

 

I have no idea how to even approach this problem. I'm thinking Pythagoras or Trig, but where do I even get the angles? From the position of the sun connected to the stick? Please help. Maybe this is a ratio type problem. 

Answer 1

Here's how to approach the problem, without giving the actual answer...

 

As the earth rotates, the angle the stick makes will change.  In fact, the angle the stick makes will be equal to the angle that the earth has rotated since noon.  Since there are 24 hours in a day, and a day is a full rotation, that means that each hour of rotation will cover 360 / 24 = 15°.

 

So At 1:45 PM, the earth will have rotated for 1.75 hours, so the stick will now be pointing at an angle of 1.75 * 15 = 26.25° away from vertical.  The stick, the shadow and the line from the top of the stick will form a right triangle, with the hypotenuse equal to the length of the stick and the side opposite the 26.25° angle equal to the length of the shadow.  You can use trigonometry to find the length of the shadow. Since you don't know the actual length of the stick, call it x and get the length of the shadow in terms of x.

 

At 2:30, the earth will have rotated for 2.5 hours, so the stick will be pointing at an angle of 37.5° away from vertical.  Again find the length of the shadow using trigonometry.  Again the length will be in terms of x, then length of the stick.

 

Now divide the shadow length at 1:45 by the length at 2:30.  The x's will cancel out, and you will be left with a decimal number that you can use to find the percent difference.