A plane is flying horizontal at a height of 1000m with a velocity of `100 ms^(-1)` when a bomb is released from it. Find (i) the time taken by it to reach the ground (ii) the velocity with which the bomb hits the target and (iii) the distance of the target.

Answer

To solve the problem step by step, we will break it down into three parts as per the requirements of the question.

### Step 1: Time taken by the bomb to reach the ground

We know that the bomb is released from a height \( h = 1000 \, \text{m} \) and it falls under the influence of gravity. The initial vertical velocity \( u_y = 0 \, \text{m/s} \) (since it is released and not thrown downwards).

Using the second equation of motion for vertical displacement:
\[ h = u_y t + \frac{1}{2} g t^2 \]
Substituting the known values:
\[ 1000 = 0 \cdot t + \frac{1}{2} \cdot 10 \cdot t^2 \]
This simplifies to:
\[ 1000 = 5 t^2 \]
Rearranging gives:
\[ t^2 = \frac{1000}{5} = 200 \]
Taking the square root:
\[ t = \sqrt{200} = 10\sqrt{2} \, \text{s} \]

### Step 2: Velocity with which the bomb hits the target

To find the final velocity \( v \) of the bomb just before it hits the ground, we can use the work-energy principle or the third equation of motion:
\[ v^2 = u_y^2 + 2gh \]
Substituting the known values:
\[ v^2 = 0 + 2 \cdot 10 \cdot 1000 \]
This simplifies to:
\[ v^2 = 20000 \]
Taking the square root:
\[ v = \sqrt{20000} = 100\sqrt{2} \, \text{m/s} \]

### Step 3: Distance of the target

The horizontal distance \( d \) traveled by the bomb can be calculated using the horizontal velocity and the time of flight. The horizontal velocity \( v_x = 100 \, \text{m/s} \) and the time of flight \( t = 10\sqrt{2} \, \text{s} \).

Using the formula:
\[ d = v_x \cdot t \]
Substituting the values:
\[ d = 100 \cdot 10\sqrt{2} \]
This simplifies to:
\[ d = 1000\sqrt{2} \, \text{m} \]

### Summary of Results
1. Time taken by the bomb to reach the ground: \( t = 10\sqrt{2} \, \text{s} \)
2. Velocity with which the bomb hits the target: \( v = 100\sqrt{2} \, \text{m/s} \)
3. Distance of the target: \( d = 1000\sqrt{2} \, \text{m} \)