Answer :
Answer:
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
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:
[tex]W=K.E\\\\Fd=\frac{1}{2}mv^2\\\\v=\sqrt{\frac{2Fd}{m}}\\\\v=\sqrt{\frac{2(100\ N)(0.5\ m)}{0.15\ kg}}[/tex]
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
Learn more about the law of conservation of energy here:
brainly.com/question/20971995?referrer=searchResults
The attached picture explains the law of conservation of energy.
