A projectile, thrown with velocity `v_(0)` at an angle `alpha` to the horizontal, has a range R. it will strike a vertical wall at a distance `R//2` from the point of projection with a speed of

Options

A. `v_(0)`
B. `v_(0)sin alpha`
C. `v_(0) cos alpha`
D. `sqrt((gR)/(2))`

Explanation

To solve the problem of a projectile striking a vertical wall at a distance \( R/2 \) from the point of projection, we can follow these steps:

### Step 1: Understand the motion of the projectile
A projectile is thrown with an initial velocity \( v_0 \) at an angle \( \alpha \) to the horizontal. The motion can be analyzed in two dimensions: horizontal and vertical.

### Step 2: Determine the range of the projectile
The range \( R \) of a projectile launched at an angle \( \alpha \) with an initial velocity \( v_0 \) is given by the formula:
\[ R = \frac{v_0^2 \sin(2\alpha)}{g} \]
where \( g \) is the acceleration due to gravity.

### Step 3: Calculate the time to reach \( R/2 \)
The horizontal distance to the wall is \( R/2 \). The horizontal component of the initial velocity is:
\[ v_{0x} = v_0 \cos(\alpha) \]
The time \( t \) taken to reach \( R/2 \) can be calculated using the formula:
\[ t = \frac{\text{distance}}{\text{velocity}} = \frac{R/2}{v_{0x}} = \frac{R/2}{v_0 \cos(\alpha)} \]

### Step 4: Find the vertical position at \( R/2 \)
The vertical position \( y \) of the projectile at time \( t \) is given by:
\[ y = v_{0y} t - \frac{1}{2} g t^2 \]
where \( v_{0y} = v_0 \sin(\alpha) \). Substituting \( t \) from the previous step:
\[ y = v_0 \sin(\alpha) \left(\frac{R/2}{v_0 \cos(\alpha)}\right) - \frac{1}{2} g \left(\frac{R/2}{v_0 \cos(\alpha)}\right)^2 \]

### Step 5: Calculate the speed of the projectile just before it strikes the wall
At the point of impact with the wall, the horizontal component of the velocity remains unchanged:
\[ v_x = v_0 \cos(\alpha) \]
The vertical component of the velocity \( v_y \) just before impact can be calculated using:
\[ v_y = v_{0y} - g t = v_0 \sin(\alpha) - g \left(\frac{R/2}{v_0 \cos(\alpha)}\right) \]

### Step 6: Calculate the resultant speed at the point of impact
The magnitude of the velocity \( v \) just before striking the wall can be found using the Pythagorean theorem:
\[ v = \sqrt{v_x^2 + v_y^2} \]
Substituting the values:
\[ v = \sqrt{(v_0 \cos(\alpha))^2 + \left(v_0 \sin(\alpha) - g \left(\frac{R/2}{v_0 \cos(\alpha)}\right)\right)^2} \]

### Step 7: Simplify the expression
After substituting \( R \) and simplifying, we find that at the point of impact, the vertical component of the velocity becomes zero at the maximum height, and the speed of the projectile when it strikes the wall is:
\[ v = v_0 \cos(\alpha) \]

### Final Answer
The speed of the projectile when it strikes the vertical wall at a distance \( R/2 \) from the point of projection is:
\[ v = v_0 \cos(\alpha) \]

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