A bomber moving horizontally with `500m//s` drops a bomb which strikes ground in 10s. The angle of strike with horizontal is

Options

A. `tan^(-1) ((1)/(5))`
B. `60^(@)`
C. `45^(@)`
D. `tan^(-1)(5)`

Explanation

To solve the problem of finding the angle of strike of a bomb dropped from a bomber moving horizontally, we can follow these steps:

### Step 1: Understand the Motion
The bomber is moving horizontally with a speed of 500 m/s. When the bomb is dropped, it has an initial horizontal velocity of 500 m/s and an initial vertical velocity of 0 m/s.

### Step 2: Identify Given Values
- Horizontal velocity (u_x) = 500 m/s
- Time of flight (t) = 10 s
- Acceleration due to gravity (g) = 10 m/s² (approximation)

### Step 3: Calculate the Vertical Velocity at Impact
To find the vertical velocity (v_y) when the bomb strikes the ground, we use the equation of motion:
\[ v_y = u_y + g \cdot t \]
Since the initial vertical velocity (u_y) is 0, we have:
\[ v_y = 0 + 10 \cdot 10 = 100 \, \text{m/s} \]

### Step 4: Calculate the Angle of Strike
The angle of strike (θ) can be calculated using the tangent function, which relates the vertical and horizontal components of velocity:
\[ \tan(\theta) = \frac{v_y}{u_x} \]
Substituting the values we found:
\[ \tan(\theta) = \frac{100}{500} = \frac{1}{5} \]

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

### Conclusion
Thus, the angle of strike with the horizontal is:
\[ \theta = \tan^{-1}\left(\frac{1}{5}\right) \]