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DerKrebs [107]
3 years ago
11

A ball encounters no air resistance when thrown into the air with 100 J of kinetic energy, which is transformed to gravitational

potential energy at the top of its trajectory. What is its kinetic energy when it returns to its original height
Physics
2 answers:
nikdorinn [45]3 years ago
4 0

Answer:

The kinetic energy when it returns to its original height is 100 J

Explanation:

The ball is thrown up with a Kinetic Energy K. E. = 0.5×m×v² = 100 J

Therefore the final height is given by

u² = v² -2·g·s

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.

finlep [7]3 years ago
4 0

Answer:

100 J

Explanation:

If there is no air resistance, then the kinetic energy is conserved. That at first the ball has 100 J of kinetic energy, when it reaches the top of its trajectory the ball has 100 J of gravitational potential energy (the kinetic energy is zero and all of them at the initial point has transformed into gravitational potential energy) , and finally when it returns to its original height it has again 100 J of kinetic energy (inverse effect than before).

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Explanation:

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RUDIKE [14]

Answer:

v = 0

Explanation:

This problem can be solved by taking into account:

- The equation for the calculation of the period in a spring-masss system

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- The equation for the velocity of a simple harmonic motion

x = \frac{2\pi }{T}Asin(\frac{2\pi }{T}t)   ( 2 )

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