Calculating Height for Half Kinetic Energy in Vertical Motion

When a body is thrown vertically upwards, its kinetic energy is gradually converted into potential energy as it gains height. At the peak of its trajectory, all the initial kinetic energy is converted into potential energy (ignoring air resistance).

In this problem, we are given the initial kinetic energy and asked to find the height at which the kinetic energy becomes half of its original value.

Given Information:

Finding the Height for Half Kinetic Energy

Let $KE_{initial}$ be the initial kinetic energy and $KE_{final}$ be the kinetic energy at the desired height $h$.

According to the problem, the kinetic energy at height $h$ is half of the original value:

$$KE_{final} = \frac{1}{2} KE_{initial}$$

Substituting the given initial kinetic energy:

$$KE_{final} = \frac{1}{2} \times 500 \text{ J} = 250 \text{ J}$$

As the body moves upwards, its kinetic energy decreases, and its potential energy increases. The total mechanical energy (sum of kinetic and potential energy) remains constant if we ignore air resistance. The loss in kinetic energy is equal to the gain in potential energy.

Loss in Kinetic Energy = $KE_{initial} - KE_{final}$

$$Loss \text{ in } KE = 500 \text{ J} - 250 \text{ J} = 250 \text{ J}$$

This loss in kinetic energy is converted into potential energy gain ($\Delta PE$). The potential energy gained when the body reaches a height $h$ is given by:

$$\Delta PE = mgh$$

Setting the loss in kinetic energy equal to the gain in potential energy:

$$Loss \text{ in } KE = \Delta PE$$

$$250 \text{ J} = mgh$$

Now, substitute the given values for mass ($m$) and acceleration due to gravity ($g$):

$$250 = (2 \text{ kg}) \times (10 \text{ m/s}^2) \times h$$

$$250 = 20h$$

To find the height $h$, divide both sides by 20:

$$h = \frac{250}{20}$$

$$h = 12.5 \text{ m}$$

Thus, the height at which the kinetic energy of the body becomes half of the original value is 12.5 m.

Revision Table: Key Concepts

Additional Information on Energy and Vertical Motion

• When a body is thrown upwards, its speed decreases as it rises, causing its kinetic energy to decrease.
• As the body gains height, its potential energy increases.
• At the maximum height, the body momentarily stops, so its kinetic energy is zero, and all the initial kinetic energy has been converted into potential energy.
• On the way down, potential energy is converted back into kinetic energy, and the body's speed increases.
• The total mechanical energy at any point during the motion (ignoring air resistance) is equal to the initial kinetic energy. Let $h$ be the height, $v$ be the velocity at that height. Total Energy = $KE + PE = \frac{1}{2}mv^2 + mgh$. This must equal the initial total energy, which is $KE_{initial}$ (assuming initial height is 0), so $\frac{1}{2}mv^2 + mgh = KE_{initial}$.
• In this problem, we used the concept that the energy lost from the kinetic form is gained in the potential form. This is a direct application of energy conservation for this scenario. Initial Energy ($h=0$) = $KE_{initial} + PE_{initial} = 500 \text{ J} + 0 \text{ J} = 500 \text{ J}$. At height $h$, Energy = $KE_{final} + PE_{final} = 250 \text{ J} + mgh$. By conservation, $500 = 250 + mgh$, which gives $mgh = 250$, leading to the same result.