A wooden block of mass 10 gm is dropped from the top of a cliff 100 m high. Simultaneously a bullet of mass 10 gm is fired from the foot of the cliff upward with a velocity 100 m/s . The bullet and the wooden will meet each other after a time in seconds

Options

A. 10
B. 0.5
C. 1
D. 7

Explanation

To solve the problem, we need to analyze the motion of both the wooden block and the bullet.

### Step 1: Understand the motion of the wooden block
The wooden block is dropped from the top of the cliff, which means its initial velocity (u) is 0 m/s. The distance (s1) it travels after time t can be calculated using the equation of motion:

\[ s_1 = ut + \frac{1}{2}gt^2 \]

Since \( u = 0 \):

\[ s_1 = 0 + \frac{1}{2}gt^2 \]
\[ s_1 = \frac{1}{2}gt^2 \]

Where:
- \( g \) = acceleration due to gravity ≈ 9.8 m/s².

### Step 2: Understand the motion of the bullet
The bullet is fired upwards with an initial velocity of 100 m/s. The distance (s2) it travels after time t can be calculated using the equation of motion:

\[ s_2 = ut + \frac{1}{2}(-g)t^2 \]

Here, the acceleration due to gravity is negative because it acts downwards:

\[ s_2 = 100t - \frac{1}{2}gt^2 \]

### Step 3: Set up the equation for the total distance
The total distance traveled by both the wooden block and the bullet must equal the height of the cliff, which is 100 m. Therefore, we can write:

\[ s_1 + s_2 = 100 \]

Substituting the expressions for s1 and s2:

\[ \frac{1}{2}gt^2 + (100t - \frac{1}{2}gt^2) = 100 \]

### Step 4: Simplify the equation
The \( \frac{1}{2}gt^2 \) terms cancel out:

\[ 100t = 100 \]

### Step 5: Solve for time t
Now, we can solve for t:

\[ t = \frac{100}{100} = 1 \text{ second} \]

### Conclusion
The bullet and the wooden block will meet each other after **1 second**.