Answer:
9.22 m/s
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.