Respuesta :
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
The horizontal speed of the soccer ball kicked horizontally off a 22 meter high hill and lands 35 meters from the edge of the hill will be V=16.5 m/s
What is Velocity?
The velocity is defined as the movement of the object with respect to the time. velocity is measured in meter per second.
[tex]y = y_0 +v_0\timest +0.5at^2[/tex]
where
[tex]y_0[/tex] = initial vertical position = 22m
y = final vertical position = 0m
[tex]v_0[/tex] = initial vertical velocity = 0 m/s
a = acceleration = -9.8 m/s^2
t = time in seconds
Now by putting the values in the formula
[tex]0 = 22 +0\times t + 0.5(-9.8)t^2[/tex]
[tex]t^2 = \dfrac{22}{4.9} = 4.49[/tex]
t = 2.12 s
So a soccer ball traveled 35m in 2.12 s
The horizontal distance traveled is determined by:
[tex]x = x_0 +v_0\timest +0.5at^2[/tex]
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
[tex]35 m = 0 +v_0\times t[/tex]
[tex]35 m = v_0\times (2.12 s)[/tex]
[tex]v_0 = 16.5 \ m/s[/tex]
So the ball was kicked 16.5 m/s in the horizontal direction
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