A baseball player exerts a force of 100 N on a ball for a distance of 0.5 mas he throws it. If the ball has a mass of 0.15 kg, what is its velocity as it leaves his hand?

This question involves the concepts of the law of conservation of energy and kinetic energy.

The baseball leaves the hand with a velocity of "25.82 m/s".

From the law of conservation of energy the work done on the ball must be equal to the kinetic energy of the ball:

where,

F = force applied = 100 N

d = displacement of ball = 0.5 m

m = mass of ball = 0.15 kg

v = 25.82 m/s

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25.82 m/s

Explanation:

We are given;

Force exerted by baseball player; F = 100 N

Distance covered by ball; d = 0.5 m

Mass of ball; m = 0.15 kg

Now, to get the velocity at which the ball leaves his hand, we will equate the work done to the kinetic energy.

We should note that work done is a measure of the energy exerted by the baseball player.

Thus;

F × d = ½mv²

100 × 0.5 = ½ × 0.15 × v²

v² = (2 × 100 × 0.5)/0.15

v² = 666.67

v = √666.67

v = 25.82 m/s

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