If a body of mass 3 kg is dropped from the top of a tower, then its kinetic energy after 3 second will be

Options

A. 1296.5J
B. 735J
C. zero
D. 1048J

Explanation

To find the kinetic energy of a body of mass 3 kg dropped from a tower after 3 seconds, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- The mass of the body (m) = 3 kg.
- The initial velocity (u) = 0 m/s (since the body is dropped).
- The acceleration due to gravity (g) = 9.8 m/s².

2. **Calculate the Velocity After 3 Seconds:**
- We use the equation of motion:
\[ v = u + gt \]
- Substitute the values:
\[ v = 0 + (9.8 \, \text{m/s}^2)(3 \, \text{s}) = 29.4 \, \text{m/s} \]

3. **Calculate the Kinetic Energy:**
- The formula for kinetic energy (KE) is:
\[ KE = \frac{1}{2} mv^2 \]
- Substitute the values of mass and velocity:
\[ KE = \frac{1}{2} (3 \, \text{kg}) (29.4 \, \text{m/s})^2 \]
- Calculate \( (29.4)^2 \):
\[ (29.4)^2 = 864.36 \, \text{m}^2/\text{s}^2 \]
- Now calculate the kinetic energy:
\[ KE = \frac{1}{2} (3) (864.36) = 1.5 \times 864.36 = 1296.54 \, \text{J} \]

4. **Final Result:**
- The kinetic energy of the body after 3 seconds is approximately **1296.54 Joules**.