any physics majors out there???

[royalk4]

Hey for my phys class we are dealing with refraction and I am convinced I have this problem correct but my HW says Im wrong.

200cm Pole 45cm out of water. Incident Angle is 55o. refracted indexes are water 1.33 and air 1.00. so I went 1.00sin(55)=1.33sin(theta2) theta2 is 38.0176. It wants to know the length of the shadow of the pole on the bottom of the pool. So amount of pole in water is 155cm. I then took tan(theta2)=d/155cm. and get 121.2cm. Hoever this apparently isn't correct??? any help?

 

[IHateMyJob2004]

T minus ....

 

[royalk4]

I tried T minus... doesn't work

 

[OOBradm]

*stares blanky at computer monitor for 1 minute then leaves thread*

 

[royalk4]

come on it's easy physics... I'm just dumb

 

[iwantanewcomputer]

is that 55 from vertical or horazontal?

 

[iwantanewcomputer]

is it 126.97?

 

[eits]

are you sure that you aren't told that the sunlight hits the water at 55 degrees? you can't answer the question without the question telling you the angle of incidence of the sunlight, because if it's past the critical point, there won't be a shadow... only a reflection.

physics was WAAAAY back in high school for me, so i gotta try and think back.

 

[eits]

if you're given that the pole is going STRAIGHT into the water and the angle of incidence of the SUNLIGHT is 55 degrees, then you take:

sin^-1 (theta) = sin55/1.33

you should get 38 degrees.

given that you have a right triangle in which the shadow is being cast at a 38 degree angle from a verticle, you know that:

a) the angles of the triangle are 38, 52, and 90
b) the long leg of the triangle (the pole) is 155 cm

then, using that, use simple math (i prefer law of sines): sin(a)/a = sin(b)/b = sin(c)/c.... so, you take:

you should come out to 96.75cm

 

[eits]

am i right?

 

[royalk4]

no the answer was 105 and change. What you needed to do was take the angle of the 1st triangle formed from the top of the pole to the water. Which was 35 since it's a right triangle and the angle entering the water is 55 still. Then do tan(35)45. 45 being the height of the pole to the water. which gives you 31 and change. Then you take the 38 and solve for x on the bottom of the pool with the same forumla and get 74.089. Add together and get 105.509. Thanks for your time and input though

 

[eits]

??? that doesn't make much sense to me.... at all.

oh, well... i guess that's why i'm getting my doctorate and not doing jack-ass-crap with physics (well, not to any mathematical extent)

 

[iwantanewcomputer]

Originally posted by: royalk4
no the answer was 105 and change. What you needed to do was take the angle of the 1st triangle formed from the top of the pole to the water. Which was 35 since it's a right triangle and the angle entering the water is 55 still. Then do tan(35)45. 45 being the height of the pole to the water. which gives you 31 and change. Then you take the 38 and solve for x on the bottom of the pool with the same forumla and get 74.089. Add together and get 105.509. Thanks for your time and input though

that's what i did, but i used 45 instead of 35...yay adding

 

[royalk4]

that was my mistake when I did it at first