Question

A soccer ball is kicked horizontally off a 24.0 meter high hill and lands a distance of 33.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.

Answer 1

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:

Answer 2

Answer:

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