A fixed mortar fires a bomb at an angle of `53^@` above the horizontal with a muzzle velocity of `80 ms^-1`. A tank is advancing directly towards the mortar on level ground at a constant speed of `5m//s`. The initial separation (at the instant mortar is fired) between the mortar and tank, so that the tank would be hit is `[Take g = 10 ms^-2]`

Options

A. 662.4m
B. 526.3m
C. 486.6m
D. None of these

Explanation

To solve the problem, we need to find the initial separation between the mortar and the tank so that the tank is hit when the bomb is fired. We will break down the solution into steps.

### Step 1: Calculate the Time of Flight of the Projectile
The time of flight (T) for a projectile launched at an angle θ with an initial velocity u is given by the formula:
\[ T = \frac{2u \sin \theta}{g} \]
Where:
- \( u = 80 \, \text{m/s} \) (muzzle velocity)
- \( \theta = 53^\circ \)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)

First, we need to calculate \( \sin 53^\circ \). Using a calculator or trigonometric tables, we find:
\[ \sin 53^\circ \approx 0.7986 \]

Now substituting the values into the formula:
\[ T = \frac{2 \times 80 \times 0.7986}{10} = \frac{127.776}{10} = 12.7776 \, \text{s} \approx 12.8 \, \text{s} \]

### Step 2: Calculate the Distance Traveled by the Tank
The tank is moving towards the mortar at a constant speed of \( 5 \, \text{m/s} \). The distance (x) the tank travels in time T is given by:
\[ x = \text{speed} \times \text{time} = 5 \, \text{m/s} \times 12.8 \, \text{s} = 64 \, \text{m} \]

### Step 3: Calculate the Range of the Projectile
The range (R) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Where:
- \( \sin 2\theta = \sin(106^\circ) \)

Using the double angle formula:
\[ \sin 2\theta = 2 \sin 53^\circ \cos 53^\circ \]
We already know \( \sin 53^\circ \approx 0.7986 \) and \( \cos 53^\circ \approx 0.6018 \).

Calculating \( \sin 2\theta \):
\[ \sin 2\theta = 2 \times 0.7986 \times 0.6018 \approx 0.9607 \]

Now substituting into the range formula:
\[ R = \frac{80^2 \times 0.9607}{10} = \frac{6400 \times 0.9607}{10} = 614.4 \, \text{m} \]

### Step 4: Calculate the Initial Separation
The initial separation (S) between the mortar and the tank when the bomb is fired is the sum of the range of the projectile and the distance traveled by the tank:
\[ S = R + x = 614.4 \, \text{m} + 64 \, \text{m} = 678.4 \, \text{m} \]

### Final Answer
The initial separation between the mortar and the tank so that the tank would be hit is approximately \( 678.4 \, \text{m} \).

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