Now consider a new situation: the ball is thrown upward from the ground with an initial velocity that takes exactly the same time tba = tab = 0.45 s to pass by the window, with the ball moving up rather than down. consider the ball’s slowdown during this time: let v ′ b be the ball’s speed (do not confuse the speed with the velocity) as it passes the window’s bottom on the way up and let v ′ a be its speed as it passes the window’s top, also in its way up. how does the ball’s slowdown ∆vup = v ′ b − v ′ a compare to its speedup ∆vdown on the way down? 1. ∆vup < ∆vdown if the mass of the ball is less than 0.1 kg and ∆vup > ∆vdown if the mass of the ball is greater than 0.1 kg 2. ∆vup < ∆vdown. 3. ∆vup = ∆vdown. 4. ∆vup > ∆vdown if the mass of the ball is less than 0.1 kg and ∆vup < ∆vdown if the mass of the ball is greater than 0.1 kg