Why Ball A travels 4 times more than Ball B?

Question

Given they are dropped at same time from different level. Ball A is twice speed of Ball B. Time reach ground is different for both balls. What should formulae used for this explanation?

Answer 1

v = -g*t (Since they start at rest, velocity is just acceleration times time)

vA = 2*vB (It's given that the velocity of ball A is twice that of B)

therefore, tA = 2*tB (A falls for twice as long)

y (distance) = .5 * t^2 (Again, since initial velocity is zero)

yA = .5*(2vB)^2 = 2 * vB^2 (Plugging in vA, which equals 2*vB, to find yA, the distance ball A travels)

yB = .5*(vB)^2

therefore, yA = 4 * yB