A ball of weight 500g is thrown in the air with a speed of 40m/s. What will be its energy at maximum height?

Options

A. 200J
B. 400J
C. 600J
D. 800J

Explanation

To solve the problem, we need to determine the energy of the ball at its maximum height after being thrown. We will follow these steps:

### Step 1: Understand the Problem
We have a ball with a mass of 500 grams (which is 0.5 kg) that is thrown upwards with an initial speed of 40 m/s. We need to find the energy of the ball at its maximum height.

### Step 2: Identify the Types of Energy
At the maximum height, the ball will have converted all its initial kinetic energy into potential energy. Therefore, we need to calculate the initial kinetic energy when the ball is thrown.

### Step 3: Calculate the Initial Kinetic Energy
The formula for kinetic energy (KE) is given by:
\[ KE = \frac{1}{2} m v^2 \]
Where:
- \( m \) is the mass of the ball (0.5 kg)
- \( v \) is the initial velocity (40 m/s)

Substituting the values into the formula:
\[ KE = \frac{1}{2} \times 0.5 \, \text{kg} \times (40 \, \text{m/s})^2 \]
\[ KE = \frac{1}{2} \times 0.5 \times 1600 \]
\[ KE = 0.25 \times 1600 \]
\[ KE = 400 \, \text{Joules} \]

### Step 4: Energy at Maximum Height
At the maximum height, all of the kinetic energy has been converted into potential energy (PE). Therefore, the potential energy at maximum height is equal to the initial kinetic energy:
\[ PE = KE = 400 \, \text{Joules} \]

### Conclusion
The energy of the ball at its maximum height is **400 Joules**.

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