We have given, the speed of the ball at the top of the window = VT and
and the speed of the ball at the bottom of the window = VB
The acceleration we know is ‘g’, g = 9.8 m/s
The me taken to cross the window, t =0.5 s and the length of the window is S = 3 m.
We have, V2-U2 = 2g(3)
also, V = U+g(0.5) i.e VB-VT = 0.5g
(VB+VT)(VB-VT) = 6g
(VB+VT)(0.5g) = 6g
VB+VT = 12 m/s,
VT-VB = -4.9 m/s
So option (A) is correct i.e VB+VT = 12 m/s