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You throw a ball of mass 190 g upward. When the ball is 2 m above the ground, headed upward (the initial state), its speed is 18 m/s. Later, when the ball is again 2 m above the ground, this time headed downward (the final state), its speed is 18 m/s. What is the change in the kinetic energy of the ball from initial to final state?

Respuesta :

Answer:

Change in Kinetic energy is zero

Explanation:

Given

Mass of Ball [tex]= 190[/tex] grams [tex]= 0.19[/tex] Kg

Speed at height of [tex]2[/tex] m [tex]= 18 \frac{m}{s}[/tex]

Kinetic energy is equal to

[tex]\frac{1}{2} mv^2[/tex]

Where M is the mass of the object and v is the velocity of the object

Change in kinetic energy [tex]= KE_2 - KE_1[/tex]

Substituting the given values we get -

Change in kinetic energy [tex]= \frac{1}{2} m_1v_1^2 - \frac{1}{2} m_2v_2^2[/tex]

Change in kinetic energy  [tex]= \frac{1}{2} 0.19*18^2 - \frac{1}{2} 0.19*18^2\\= 30.78 - 30.78\\= 0[/tex]

Hence, the change in Kinetic energy is zero

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