The velocity and direction of projection of a shot which passes in horizontal direction just over the top of a wall which is 50 yds. away and 75 feet high.

Options

A. `40, 30^(@)`
B. `40sqrt(6), 45^(@)`
C. `50, 60^(@)`
D. None

Answer

To solve the problem of finding the velocity and direction of projection of a shot that passes just over the top of a wall, we will follow these steps:

### Step 1: Understand the Problem
We need to find the initial velocity (U) and the angle of projection (α) for a projectile that just clears a wall that is 75 feet high and located 50 yards away.

### Step 2: Convert Units
Since the height is given in feet and the distance in yards, we need to convert everything to the same unit.
1 yard = 3 feet, so:
- Distance to the wall = 50 yards = 50 × 3 = 150 feet.
- Height of the wall = 75 feet.

### Step 3: Use the Maximum Height Formula
The maximum height (H) reached by a projectile is given by the formula:
\[ H = \frac{U^2 \sin^2 \alpha}{2g} \]
Where:
- \( g \) = acceleration due to gravity = 32 ft/s² (approximately).

Substituting the values we have:
\[ 75 = \frac{U^2 \sin^2 \alpha}{2 \times 32} \]
This simplifies to:
\[ U^2 \sin^2 \alpha = 75 \times 64 \]
\[ U^2 \sin^2 \alpha = 4800 \] (Equation 1)

### Step 4: Use the Range Formula
The range (R) of a projectile is given by:
\[ R = \frac{U^2 \sin 2\alpha}{g} \]
Since the range to the wall is 150 feet, we can write:
\[ 150 = \frac{U^2 \sin 2\alpha}{32} \]
Using the identity \( \sin 2\alpha = 2 \sin \alpha \cos \alpha \), we get:
\[ 150 = \frac{U^2 (2 \sin \alpha \cos \alpha)}{32} \]
This simplifies to:
\[ U^2 \sin 2\alpha = 150 \times 32 \]
\[ U^2 \sin 2\alpha = 4800 \] (Equation 2)

### Step 5: Relate the Two Equations
From Equation 1 and Equation 2, we have:
\[ U^2 \sin^2 \alpha = U^2 \sin 2\alpha \]
This implies:
\[ \sin^2 \alpha = \sin 2\alpha \]
Using the identity \( \sin 2\alpha = 2 \sin \alpha \cos \alpha \):
\[ \sin^2 \alpha = 2 \sin \alpha \cos \alpha \]
Dividing both sides by \( \sin \alpha \) (assuming \( \sin \alpha \neq 0 \)):
\[ \sin \alpha = 2 \cos \alpha \]
This gives:
\[ \tan \alpha = 2 \]

### Step 6: Find the Angle of Projection
From \( \tan \alpha = 2 \), we can find \( \alpha \):
\[ \alpha = \tan^{-1}(2) \]
Calculating this gives approximately \( \alpha \approx 63.43^\circ \).

### Step 7: Calculate the Initial Velocity
Now substituting \( \sin \alpha \) and \( \cos \alpha \) back into Equation 1 to find \( U \):
Using \( \sin \alpha = \frac{2}{\sqrt{5}} \) and \( \cos \alpha = \frac{1}{\sqrt{5}} \):
\[ U^2 \left(\frac{2}{\sqrt{5}}\right)^2 = 4800 \]
\[ U^2 \cdot \frac{4}{5} = 4800 \]
\[ U^2 = 4800 \cdot \frac{5}{4} \]
\[ U^2 = 6000 \]
\[ U = \sqrt{6000} = 10\sqrt{60} \]

### Final Answer
The initial velocity \( U \) is \( 10\sqrt{60} \) feet per second, and the angle of projection \( \alpha \) is approximately \( 63.43^\circ \).