Asoccer 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

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