A ball is dropped from the top of a building. The ball takes 0.5 s to fall past the length of a window, the top of the window being at a distance of 3 m from the top of the building. If the speed of the ball at the top and the bottom of the window are Vₜ and Vᵦ respectively, then (g=9.8m/sec²)
a. Vₜ+Vᵦ=20ms⁻¹
b. Vₜ−Vᵦ=4.9ms⁻¹
c. VᵦVₜ=1ms⁻¹
d. Vᵦ/Vₜ=1ms⁻¹