Answer:
9.22 m/s
Step-by-step explanation:
Vertical height, y = 36 m 
Horizontal distance, d = 25 m 
Let the initial velocity of projection is u and it is projected horizontally. 
Let t be the time taken by the ball to hit the ground. 
For horizontal motion 
horizontal distance = horizontal velocity x time 
d = u x t 
25 = u x t .... (1) 
Use second equation of motion for vertical direction 
y = 0.5 x g t², as the initial vertical velocity is zero. 
36 = 0.5 x 9.8 x t² 
t = 2.7 second 
Put in equation (1) 
25 = u x 2.7 
u = 9.22 m/s 
Thus, the initial velocity with which the ball is projected horizontally is 9.22 m/s.