A shot is fired from a point at a distance of `200 m` from the foot of a tower `100 m` high so that it just passes over it horizontally. The direction of shot with horizontal is.

Options

A. `30^@`
B. `45^@`
C. `60^@`
D. `70^@`

Explanation

To solve the problem, we need to determine the angle at which the shot is fired so that it just passes over the tower horizontally. Here’s a step-by-step solution:

### Step 1: Understand the problem
We have a tower that is 100 m high and a point from which a shot is fired at a distance of 200 m from the base of the tower. The shot must just clear the top of the tower horizontally.

### Step 2: Identify the maximum height formula
In projectile motion, the maximum height (H) reached by a projectile is given by the formula:
\[ H = \frac{u_y^2}{2g} \]
where \(u_y\) is the vertical component of the initial velocity and \(g\) is the acceleration due to gravity (approximately \(10 \, \text{m/s}^2\)).

### Step 3: Set up the equation for maximum height
Since the tower is 100 m high, we set \(H = 100\) m:
\[ 100 = \frac{u_y^2}{2 \cdot 10} \]
This simplifies to:
\[ u_y^2 = 2000 \]
Taking the square root gives:
\[ u_y = \sqrt{2000} = 20\sqrt{5} \, \text{m/s} \]

### Step 4: Calculate the time to reach the maximum height
The time to reach the maximum height (t) can be calculated using the formula:
\[ t = \frac{u_y}{g} \]
Substituting the values:
\[ t = \frac{20\sqrt{5}}{10} = 2\sqrt{5} \, \text{s} \]

### Step 5: Calculate the horizontal component of the velocity
The horizontal distance covered during this time is given as 200 m. The horizontal component of the initial velocity (\(u_x\)) can be calculated using:
\[ \text{Distance} = u_x \cdot t \]
Substituting the values:
\[ 200 = u_x \cdot 2\sqrt{5} \]
Solving for \(u_x\):
\[ u_x = \frac{200}{2\sqrt{5}} = \frac{100}{\sqrt{5}} = 20\sqrt{5} \, \text{m/s} \]

### Step 6: Calculate the angle of projection
Now we have both components of the initial velocity:
- \(u_y = 20\sqrt{5}\)
- \(u_x = 20\sqrt{5}\)

The angle of projection (\(\theta\)) can be found using:
\[ \tan \theta = \frac{u_y}{u_x} \]
Substituting the values:
\[ \tan \theta = \frac{20\sqrt{5}}{20\sqrt{5}} = 1 \]
Thus,
\[ \theta = \tan^{-1}(1) = 45^\circ \]

### Final Answer:
The direction of the shot with respect to the horizontal is \(45^\circ\).

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