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. `tan^(-1)((1)/(2))`
C. `tan^(-1) (1)`
D. `tan^(-1)(5)`

Explanation

To solve the problem of finding the angle of strike of the bomb with the horizontal, we can follow these steps:

### Step 1: Identify the horizontal and vertical components of motion
The bomber is moving horizontally with a speed of 500 m/s. When the bomb is dropped, it retains this horizontal velocity. Thus, the horizontal component of velocity (Vx) is:
- Vx = 500 m/s

The bomb is dropped, so its initial vertical velocity (Uy) is:
- Uy = 0 m/s

### Step 2: Calculate the vertical component of velocity (Vy) just before impact
The bomb falls under the influence of gravity. The vertical acceleration (A) is equal to the acceleration due to gravity, which is approximately 10 m/s². The time (t) it takes for the bomb to hit the ground is given as 10 seconds.

Using the formula for final velocity:
\[ Vy = Uy + At \]
Substituting the known values:
\[ Vy = 0 + (10 \, \text{m/s}^2)(10 \, \text{s}) \]
\[ Vy = 100 \, \text{m/s} \]

### Step 3: Calculate the angle of strike (θ) with the horizontal
The angle of strike can be found using the tangent function, which relates the vertical and horizontal components of velocity:
\[ \tan(\theta) = \frac{Vy}{Vx} \]
Substituting the values we calculated:
\[ \tan(\theta) = \frac{100 \, \text{m/s}}{500 \, \text{m/s}} \]
\[ \tan(\theta) = \frac{1}{5} \]

### Step 4: Calculate the angle θ
To find the angle θ, we take the arctangent (inverse tangent) of the ratio:
\[ \theta = \tan^{-1}\left(\frac{1}{5}\right) \]

### Final Step: Conclusion
Using a calculator, we can find:
\[ \theta \approx 11.31^\circ \]

Thus, the angle of strike with the horizontal is approximately **11.31 degrees**.

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