A soccer ball is kicked horizontally off a 22m hill and lands a distance of 35 m from the edge of the hill?

Question

What is the initial velocity of the soccer ball.

I need help with Vectors can someone help me with them

Answer 1

well, since no force acts on the soccer ball in the horizontal direction once it is kicked...its horizontal velocity remains constant, and its initial vertical velocity is 0. But once the ball is kicked off the hill it has a constant acceleration in the vertical direction by gravity.

the vertical distance travelled (upward is positive) is determined by:
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

Answer 2

Assuming no drag forces, since you gave nothing to indicate them, PE = mgh = 1/2 mv^2 = KE; where PE is potential energy, KE is kinetic energy, m is the mass of the ball, g ~ m/sec^2 on Earth's surface, h = 22 m hill height, and v is the initial velocity you're looking for. Thus,

v = sqrt(2gh); you can do the math.

Lesson learned, potential energy becomes kinetic energy when a mass is dropped from a height. The horizontal forces, right after the ball leaves the foot where that "initial velocity" is, are zero if there are no drag forces.

So the only energies in this problem are those from the ball's PE based on its vertical height above the impact level. In sum, we don't care that the ball landed 35 m from the hill. That's why PE = KE in this case. A more realistic problem would fold in drag forces; in which case, we would care about how far the ball gets from the hill because of the work function W = Fd and F would be the friction force of drag and d would be = 35 m, the distance from the hill.

Answer 3

because the ball is kicked horizonally, its intial vertical velocity is 0m/s. As it falls downward, it falls at the rate of -9.8m/s^2 and also travels horizonally at a constant velocity.

find the time it takes the ball the hit the ground.
Xf = 1/2at^2 + vt + Xo

again, since the ball falls from rest, its vertical veloicty is 0m/s, we don't need vt part in the equation above.

Xf = final position, which is 0m
Xo = intial position, which is 22m

0 = 1/2(-9.8)t^2 + 22
-22 = 1/2(-9.8)t^2
-44 = -9.8t^2
t^2 = 44/9.8
t = 2.12s

the time it takes the ball to hit the ground is also the time it takes the ball to land 35m away from the hill.

x = vt
35m = v(2.12s)
v = 16.51m/s

Answer 4

Ignoring air resistance and assuming the ball was kicked horizontally, the time it takes to reach the ground is
t = √(2h/g)
where
g = acceleration due to gravity
h = height
Initial velocity
v = s/t = s/√(2h/g)
where
s = horizontal distance
v = 35/√(2(22)/9.80665)
v = 16.523 m/s