The correct answer to this question is D
Here is my step-by-step-work. Let me know if you have any questions! :)
When a mirror is rotated . . .
-- The incident ray doesn't turn. It's just the line from the source to the mirror.
It would be there, in the same place, even if there was no mirror.
-- The normal turns. It's the line perpendicular to the mirror, so it must turn
with the mirror.
-- Since the normal tuns and the incident ray doesn't, the angle between them
must change. And since the angle of the reflected ray is equal to the angle of
the incident ray, the reflected ray must also turn.
I'm not sure if this is correct but it's what I'll do
This is free-fall problem.
Stone A is thrown upward, at the point it falls down to the place where it was thrown, the velocity is -15m/s.
Now I choose the bridge is the origin. From the bridge, stone A and B fall the same distance which means Ya = Yb ( vertical distance )
Ya = Vo(t + 2) + 1/2a(t+2)^2
= -15(t + 2) + 1/2(9.8)(t^2 + 4t + 4)
= -15t - 30 + 4.5(t^2 + 4t + 4)
= -15t - 30 + 4.5t^2 + 18t + 18
= 4.5t^2 +3t - 12
Yb = Vo(t) + 1/2a(t)^2
= 0 + 4.5t^2
4.5t^2 = 4.5t^2 +3t - 12
0 = 3t - 12
4 = t
Time for Stone B is 4s
Time for Stone A is 6s
