The kinetic energy when it returns to its original height is 100 J
Solution:
The ball is thrown up with a Kinetic Energy K. E. = 0.5×m×v² = 100 J
Therefore the final height is given by
<u>u² = v² -2·g·s</u>
Where:
u = final velocity = 0
v = initial velocity
s = final height
Therefore v² = 2·g·s = 19.62·s
P.E = Potential Energy = m·g·s
Since v² = 2·g·s
Substituting the value of v² in the kinetic energy formula, we obtain
K. E. = 0.5×m×2·g·s = m·g·s = P.E. = 100 J
When the ball returns to the original height, we have
v² = u² + 2·g·s
Since u = 0 = initial velocity in this case we have
v² = 2·g·s and the Kinetic energy = 0.5·m·v²
Since m and s are the same then 0.5·m·v² = 100 J.
As the height of the ball increases the kinetic energy of the ball is converted into gravitational potential energy. This means that the kinetic energy of the bullet is reduced. When the ball reaches its maximum height, it momentarily comes to rest and the ball's kinetic energy is zero. When the ball hits the ground, its potential energy is converted to kinetic energy.
Learn more about Kinetic energy here:-brainly.com/question/25959744
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