Answer:
The one falling from the greatest height will have the greatest speed.
h = 1/2 g t^2 time for ball to fall distance h
h2 / h1 = t2^2 / t1^2 dividing equations
h2 / t2^2 = h1 / t1^2
Let v be the average speed (v2 = h2 / t2)
1 / t2 * v2 = 1 / t1 * v1
v2 / v1 = t2 / t1 the one taking the longest to fall has the greater av. speed
Check:
h4 / h1 = t4^2 / t1^2 or
t4 / t1 = (h4 / h1)^1/2
In this case t4 / t1 = (4 / 1)^1/2 = 2 or twice the average speed
t1 = (2 h / g)^1/2 = .2^1/2 = .447 using g = 10
t4 = (2 h / g)^1/2 = .8^1/2 = .894
v1 = 1 / .447 = 2.24 m/s average speed
v4 = 4 / .894 = 4.47 or twice the average speed
