A soccer ball is kicked horizontally off a bridge with a height of 36m. The ball travels 25m horizontally before it hits the pavement below. What was the soccer ball's speed when it was first kicked?

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.

15.3 cos 65 degrees trig