1.

A soccer ball is kicked horizontally off a 40.0 meter high hill and lands at a distance of 32.0 meters from the edge of the hill. The fall took 2.86 s. Determine the initial horizontal velocity of the soccer ball.

Answer»

ong>Explanation:

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

=

d

i

s

t

a

n

c

e

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

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