Aball is released from a tower at a height of 100 meters toward the roof of another tower that is 25 meters high. the horizontal distance between the two towers is 20 meters. with what horizontal velocity should the ball be imparted so that it lands on the rooftop of the second building?

Let's figure out how long it will take to fall 75 meters, from the 1st roof to the second roof. We can assume that gravity is the only force affecting vertical velocity, so the ball starts from rest and accelerates downward at g, or -9.8m/s².

Let's find how long it takes to fall 75 meters:
s(t) = Vi + (1/2)*a*t²,
where s(t) is displacement as a function of time, Vi is initial velocity (zero), and a is acceleration. Plugging in our values:
-75 = 0 + (1/2)(-9.8)(t²)      Multiply both sides by 2/-9.8
15.3 = t²                             Take the square root of both sides
t = 3.91

We need to the ball to travel 20 meters horizontally before it hits the roof in 3.91 seconds. We can assume that the horizontal velocity remains constant (a=0, Vi=V(t) for all t). 
Therefore, the minimum horizontal velocity is:
D = V*t , simple distance formula, distance equals velocity times time:
20 = V * 3.91       Divide both sides by 3.91
V = 5.11

The horizontal velocity, therefore, must be at least 5.11m/s in order for the ball to reach the roof of the second building. 

Distance^2= yellow line ^2 (see picture) = 30^2 + 49^2so distance = squareroot of (300^2 +49^2)i dont know if this is how to solve it, but i cnt think of another way