Question

A ball is thrown upward . at what height it's kinetic energy will be twice of it's potential energy?

Answer 1

Answer on Question 50658, Physics, Mechanics | Kinematics | Dynamics

Question:

A ball is thrown upward. At what height its kinetic energy will be twice of its potential energy?

Solution:

By the definition of the law of conservation of energy, when the ball is just begin to move upward we have:

$$E = P E + K E = 0 + \frac {1}{2} m v _ {0} ^ {2},$$

where $m$ is the mass of the ball, $v_{0}$ is the initial velocity of the ball.

From the condition of the question we know that when the ball reaches a certain height its kinetic energy will be twice of its potential energy:

$$\frac {1}{2} m v ^ {2} = 2 m g h.$$

Therefore, we can find the height from the law of conservation of energy:

$$0 + \frac {1}{2} m v _ {0} ^ {2} = m g h + \frac {1}{2} m v ^ {2},$$ $$\frac {1}{2} m v _ {0} ^ {2} = m g h + 2 m g h,$$ $$\frac {1}{2} m v _ {0} ^ {2} = 3 m g h,$$

$h = \frac{v_0^2}{6g} = \frac{1}{3}\frac{v_0^2}{2g} = \frac{1}{3} H$, where $H = \frac{v_0^2}{2g}$ is the maximum height reached by the ball.

Answer:

$$h = \frac {1}{3} H.$$

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