Not Motion Again!

Let the distance from the top be $x$

Then $x=ut+\frac{1}{2}gt^2 \\ \rightarrow x= 0 + 4.9t^2$

The distance from the ground will be $100-x=ut-\frac{1}{2}gt^2 \\ \rightarrow 100-x=25t-4.9t^2 \\ \rightarrow x = 100-25t+4.9t^2$

Equating the two, we get

$4.9t^2 = 100-25t+4.9t^2 \\ \rightarrow 0=100-25t \\ \rightarrow 25t=100 \\ \rightarrow t=\frac{100}{25} = \boxed {4}$

Now, we know that the time taken is $4$ seconds, so applying the value of $t$ in

$x=4.9t^2$ , we get $4.9 \times 4^2= 4.9 \times 16=78.4$

Thus the distance from the bottom is $100-78.4=21.6$

Thus, the answer is $4 + 21.6 = \huge \boxed {25.6}$

Let the two balls meet after t seconds

then distance travel by each ball in t sec :

for ball 1:

$S1=ut+\dfrac {1}{2}gt^2$

$S1= 4.9t^2$ (since ball is dropped its initial velocity $u=0$ )

for ball 2:

$S2=25t+\dfrac 12 (-9.8) t^2$ (-ve of g shows g is working against the motion)

$S2= 25t-4.9 t^2$

$S1+S2 = 100$ (height of tower)

so $t=4$

now put t=4 in S2 it will give $S2=21.6$

answer = $t+S2= \boxed{25.6}$