a soccer ball is kicked horizontally off a 22 meter high hill and lands 35 meters from the edge of the hill. determine the initial horizontal velocity of the soccer ball

y = y0 + v0*t + 0.5at^2

where y0 = initial vertical position = 22m

y = final vertical position = 0m

v0 = initial vertical velocity = 0 m/s

a = acceleration = - 9.8 m/s^2

t = time in seconds

0 = 22 + 0*t + 0.5 (-9.8) t^2

t^2 = 22/4.9 = 4.49 s^2

t = 2.12 s

So it traveled 35m in 2.12 s

the horizontal distance traveled is determined by:

x = x0 + v0*t + 0.5at^2

but here a in the horizontal direction is 0 m/s^2

and v0 is in the velocity in the horizontal direction in this equation

35 m = 0 + v0*t

35 m = v0 (2.12 s)

v0 = 16.5 m/s

So the ball was kicked 16.5 m/s in the horizontal direction