Physics Induced EMF exam question help

box A conducting rod is held horizontally in an east–west direction.
The magnetic flux density of the Earth’s magnetic field is 4.9 × 10−5 T and is directed
at an angle of 68° to the ground.
0 3 . 1 Figure 5 shows the arrangement. The rod has a length of 2.0 m.
Figure 5
The rod is released and falls 8.0 m to the ground. It remains in a horizontal
east–west direction as it falls.
Determine the average emf across the rod during its fall to the ground.
Assume that air resistance is negligible.
[3 marks]
So the diagram shows the rod falling vertically down and the magnetic field is at an angle of 68 degrees. So with the values given i know it has to be e=blvcos(68). however I do not understand why we use the final velocity before it hits the ground as v in the emf equation. Because in the question it states average emf, but if we use the max value of velocity would that not give us the max emf.
Note : the correct answer is 4.5x10^-4v, which is obtained by using the max/final velocity v which is 12.5

I agree with you
I got 2.70V

I did it a couppe of days ago and was super confused too. So ur supposed to use suvat to fin the time taken to reach ground and the calculate average speed. Divide distance by time taken for v

Really because when i put it into my calculator i get the correct answer. Are u perhaps on radians mode instead of degrees?

Okay so E=Blv, i multiplied together: 2 x cos68 x 6.26 x 4.9x10^-5 and that gives me 2.3x10^-4. My v=6.26 which u get by dividing 8 by 1.27

?? Kind of baffled rn cuz my answer is what the markscheme is saying is correct. 😭😭 are u looking at the official one?

Yep, my teacher sent it to me and he got it from the aqa website. Where is your mark scheme from?