Answer:
Question:
A soccer ball is kicked horizontally off a
22.0
meter high hill and lands a distance of
35.0
meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.
Projectile:
The projectile motion is a combination of uniform motion and uniform acceleration motion. When an object is thrown with a velocity (v) then, the horizontal component of the velocity (
v
h
) does not change with respect to time and therefore, the following relation holds valid for the horizontal motion:
S
p
e
e
d
=
S
p
e
e
d
T
i
m
e
The vertical component of the velocity (
v
v
) changes at the rate of
g
=
9.81
m
/
s
2
and therefore, the equations of motion are used to study the vertical motion of the object. These equations are customized for the vertical motion as:
v
=
u
+
g
t
s
=
u
t
+
0.5
g
t
2
v
2
−
u
2
=
2
g
s
For an object thrown with a horizontal velocity, the initial vertical component of the velocity will be zero. In that case, the vertical distance traveled by the object in time (t) will be,
s
=
0.5
g
t
2
⇒
t
=
√
2
s
g
Answer and Explanation:
Given:
Height of the hill,
H
=
22
m
The horizontal distance covered by the ball before it lands,
d
=
35
m
Calculating the time taken by the ball to hit the ground,
t
=
√
2
H
g
=
√
2
(
22
)
9.81
=
2.12
s
Calculating the initial horizontal velocity of the ball,
V
=
d
t
=
35
2.12
=
16.51
m
/
s
which is the answer.
Explanation:
