If a body of mass 3 kg is droped from the top of a tower of height 25 metre, then its kinetic energy after 1 second will be`(g = 10 m/s^(2))`

Options

A. `1126 J`
B. `1048 J`
C. `735 J`
D. `150 J`

Explanation

To find the kinetic energy of a body after 1 second of being dropped from a height, we can follow these steps:

### Step 1: Identify the given values
- Mass of the body (m) = 3 kg
- Height of the tower (h) = 25 m (not directly needed for this calculation)
- Acceleration due to gravity (g) = 10 m/s²
- Time (t) = 1 s

### Step 2: Calculate the final velocity after 1 second
Using the first equation of motion:
\[ v = u + gt \]
where:
- \( u \) = initial velocity = 0 (since the body is dropped)
- \( g \) = acceleration due to gravity = 10 m/s²
- \( t \) = time = 1 s

Substituting the values:
\[ v = 0 + (10 \, \text{m/s}^2)(1 \, \text{s}) \]
\[ v = 10 \, \text{m/s} \]

### Step 3: Calculate the kinetic energy
The formula for kinetic energy (KE) is:
\[ KE = \frac{1}{2} mv^2 \]
Substituting the values of mass and velocity:
\[ KE = \frac{1}{2} \times 3 \, \text{kg} \times (10 \, \text{m/s})^2 \]
\[ KE = \frac{1}{2} \times 3 \times 100 \]
\[ KE = \frac{300}{2} \]
\[ KE = 150 \, \text{Joules} \]

### Final Answer
The kinetic energy of the body after 1 second is **150 Joules**.

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