A bomber plane moves horizontally with a speed of `500 m//s` and a bomb released from it, strikes the ground in `10 s`. The angle with horizontally at which it strikes the ground will be

Options

A. `tan^(-1) (1/5)`
B. `tan^(-1) (1/15)`
C. `tan^(-1) (1)`
D. `tan^(-1) (5)`

Answer

To solve the problem, we need to determine the angle at which the bomb strikes the ground after being released from a bomber plane moving horizontally. Here’s a step-by-step solution:

### Step 1: Understand the motion of the bomb
When the bomb is released from the plane, it has an initial horizontal velocity equal to that of the plane, which is \(500 \, \text{m/s}\). The vertical motion is influenced by gravity.

### Step 2: Analyze vertical motion
The bomb is in free fall, so we can calculate its vertical velocity after \(10 \, \text{s}\). The initial vertical velocity (\(u_y\)) is \(0 \, \text{m/s}\) since it is released from the plane. The acceleration due to gravity (\(a_y\)) is \(g = 10 \, \text{m/s}^2\) (assuming this value for simplicity).

Using the equation of motion:
\[ v_y = u_y + a_y \cdot t \]
Substituting the values:
\[ v_y = 0 + 10 \cdot 10 = 100 \, \text{m/s} \]
So, the vertical velocity (\(v_y\)) just before it strikes the ground is \(100 \, \text{m/s}\).

### Step 3: Identify horizontal motion
The horizontal velocity (\(u_x\)) remains constant throughout the motion since there is no horizontal acceleration. Thus:
\[ u_x = 500 \, \text{m/s} \]

### Step 4: Calculate the angle of impact
We need to find the angle \(\theta\) with the horizontal at which the bomb strikes the ground. We can use the tangent function:
\[ \tan(\theta) = \frac{v_y}{u_x} \]
Substituting the known values:
\[ \tan(\theta) = \frac{100}{500} = \frac{1}{5} \]

### Step 5: Find the angle \(\theta\)
To find \(\theta\), we take the arctangent:
\[ \theta = \tan^{-1}\left(\frac{1}{5}\right) \]

### Final Answer
The angle with the horizontal at which the bomb strikes the ground is:
\[ \theta = \tan^{-1}\left(\frac{1}{5}\right) \]