There are no guesses in this problem. Basically, there are two instances to solve the unknown.... one is when is the initial point of reference wherein time t =0, and the other is the time when the two balls are at the same height which we need to compute. At this point, t = unknown.
Usinfg the equation for ball A, we could solve this since we know all the initial condition for A. So,
to solve for the the time when the dist travelled by ball A is 20 w/ ref to ground:
20 = -½gt² + 40, xa ; 20 = (-1/2)gt^2 ; 40 = gt²
t = 1.114556425s
So now, we could use this info to get the initial velocity of ball B. Plugging in the values: using the formula, x = xb + Vb*t+ -½gt², wherein xb is the initial position of ball b and Vb is the initial velocity of ball B.
At time t = 1.114556425s, x = 20ft from the given.
20 = 5 + Vb*t + -½gt², where -½gt² = 20 from the initial computation.
we get,
20 = 5 - 20 + Vb*t
35 = Vb*t
Vb = 35/1.114556425 = 31.402627 ft/s!
So, there are no guesses here.
going back to your other questions,
If it is stated that the ball is thrown upward or downward, it means that the ball has some initial velicty. If its released, thats the point we could say that its free fall meaning, the initial velicity is zero.
for the last question: How to obtain the following Equation on "Given"? h3 = –½gt² + h1 ,
this was derived for the ball A from the satndard equation,
x = xa + Va*t + ½at².... where xa = init position = 40;Va = init velocity = 0 ( dropped!); a= acceleration = -g; and final position = 20
so it was simplified to
h3 = –½gt² + h1
Hope this helps.