Energy conservation

. A pool ball of mass m is shot off a 0.68 m high pool table and hits the floor with a speed

of 6.0 m/s. What was the ball’s initial velocity?

Respuesta :

Answer:

7.328m/s

Explanation:

Given parameters:

height of table = 0.68m

final velocity of the ball = 6m/s

Unknown:

Initial velocity of ball = ?

Solution:

To solve this problem, we are going to employ the appropriate motion equation.

We must understand that this fall occurs in the presence of gravity;

            V = U + 2gH

Where;

 V is the final velocity

 U is the initial velocity

 g is the acceleration due to gravity

 H is the height of the pool table

    Since U is the unknown, let us make it the subject of the expression;

           U = V - 2gH

        U = 6 - (2 x 9.8 x 0.68)  = 7.328m/s(deceleration)

The ball's initial velocity will be "7.328 m/s (deceleration)".

According to the question,

Height,

  • h = 0.68 m

Final velocity,

  • v = 6 m/s

We know the relation,

→ [tex]v = u +2 gh[/tex]

or,

→ [tex]u = v-2gh[/tex]

By putting the values, we get

      [tex]= 6-(2\times 9.8\times 0.68)[/tex]

      [tex]= 6 -13.328[/tex]

      [tex]= 7.328 \ m/s[/tex] (deceleration)

Thus the approach above is right.

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