A shot fired from a gun on top of a tower, 272 feet high hits the ground at a distance of 4352 feet in 17 seconds. The velocity and direction of projection are

Options

A. `256, 30^(@)`
B. `256sqrt(2), 45^(@)`
C. `180^(@), 60^(@)`
D. None

Answer

To solve the problem of a shot fired from a gun on top of a tower, we need to find the velocity and direction of projection. Here’s a step-by-step solution:

### Step 1: Identify the Given Data
- Height of the tower (h) = 272 feet
- Horizontal distance (range, r) = 4352 feet
- Time of flight (t) = 17 seconds
- Acceleration due to gravity (g) = 32 ft/s² (approximately, since we are using feet)

### Step 2: Set Up the Vertical Motion Equation
Using the second equation of motion for vertical displacement:
\[ y = ut \sin \theta - \frac{1}{2} g t^2 \]
Here, the vertical displacement (y) is -272 feet (since it falls down), so we have:
\[ -272 = u \sin \theta \cdot 17 - \frac{1}{2} \cdot 32 \cdot (17)^2 \]

### Step 3: Calculate the Vertical Component
Substituting the values into the equation:
\[ -272 = 17u \sin \theta - \frac{1}{2} \cdot 32 \cdot 289 \]
Calculating \(\frac{1}{2} \cdot 32 \cdot 289\):
\[ \frac{1}{2} \cdot 32 \cdot 289 = 16 \cdot 289 = 4624 \]
So, we have:
\[ -272 = 17u \sin \theta - 4624 \]
Rearranging gives:
\[ 17u \sin \theta = 4624 - 272 \]
\[ 17u \sin \theta = 4352 \]
Thus:
\[ u \sin \theta = \frac{4352}{17} = 256 \]

### Step 4: Set Up the Horizontal Motion Equation
For horizontal motion, since there is no acceleration:
\[ r = u \cos \theta \cdot t \]
Substituting the values:
\[ 4352 = u \cos \theta \cdot 17 \]
Rearranging gives:
\[ u \cos \theta = \frac{4352}{17} = 256 \]

### Step 5: Relate the Two Components
Now we have two equations:
1. \(u \sin \theta = 256\)
2. \(u \cos \theta = 256\)

Dividing the first equation by the second:
\[ \frac{u \sin \theta}{u \cos \theta} = \frac{256}{256} \]
This simplifies to:
\[ \tan \theta = 1 \]
Thus:
\[ \theta = 45^\circ \]

### Step 6: Calculate the Initial Velocity (u)
Using either equation, we can find \(u\):
\[ u \sin 45^\circ = 256 \implies u \cdot \frac{\sqrt{2}}{2} = 256 \]
Thus:
\[ u = 256 \cdot \frac{2}{\sqrt{2}} = 256 \sqrt{2} \]

### Final Result
The initial velocity \(u\) is \(256\sqrt{2}\) feet per second, and the angle of projection \(\theta\) is \(45^\circ\).

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