A ball is projected at an angle `45^(@)` with horizontal. It passes through a wall of height `h` at horizontal distance `d_(1)` from the point of projection and strikes the ground at a horizontal distance `(d_(1) + d_(2))` from the point of projection, then

Options

A. `h = (2 d_d_(2))/(d_+ d_(2))`
B. `h = (d_(1) d_(2))/(d_(1) + d_(2))`
C. `h = (sqrt(2) d_(1) d_(2))/(d_(1) + d_(2))`
D. `h = ( d_(1) d_(2))/(2(d_(1) + d_(2)))`

Explanation

To solve the problem step by step, we need to analyze the projectile motion of the ball that is projected at an angle of \( 45^\circ \) and passes through a wall of height \( h \) at a horizontal distance \( d_1 \) before hitting the ground at a distance \( d_1 + d_2 \).

### Step 1: Understand the motion of the projectile
The ball is projected at an angle of \( 45^\circ \) with the horizontal. The horizontal and vertical components of the initial velocity \( u \) can be expressed as:
- \( u_x = u \cos(45^\circ) = \frac{u}{\sqrt{2}} \)
- \( u_y = u \sin(45^\circ) = \frac{u}{\sqrt{2}} \)

### Step 2: Determine the time to reach the wall
The horizontal distance to the wall is \( d_1 \). The time \( t_1 \) taken to reach the wall can be calculated using the horizontal motion equation:
\[ d_1 = u_x \cdot t_1 \implies t_1 = \frac{d_1}{u_x} = \frac{d_1 \sqrt{2}}{u} \]

### Step 3: Calculate the height of the ball when it reaches the wall
The vertical position \( y \) of the ball when it reaches the wall can be calculated using the vertical motion equation:
\[ y = u_y \cdot t_1 - \frac{1}{2} g t_1^2 \]
Substituting \( u_y \) and \( t_1 \):
\[ h = \frac{u}{\sqrt{2}} \cdot \frac{d_1 \sqrt{2}}{u} - \frac{1}{2} g \left(\frac{d_1 \sqrt{2}}{u}\right)^2 \]
This simplifies to:
\[ h = d_1 - \frac{g d_1^2}{2u^2} \]

### Step 4: Determine the total range of the projectile
The total horizontal distance the ball travels before hitting the ground is \( d_1 + d_2 \). The range \( R \) of a projectile launched at an angle \( \theta \) is given by:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
For \( \theta = 45^\circ \):
\[ R = \frac{u^2}{g} \]
Thus, we can set:
\[ d_1 + d_2 = \frac{u^2}{g} \]

### Step 5: Relate \( u^2 \) to \( d_1 \) and \( d_2 \)
From the equation \( d_1 + d_2 = \frac{u^2}{g} \), we can express \( u^2 \):
\[ u^2 = g(d_1 + d_2) \]

### Step 6: Substitute \( u^2 \) back into the height equation
Substituting \( u^2 \) into the height equation:
\[ h = d_1 - \frac{g d_1^2}{2(g(d_1 + d_2))} \]
This simplifies to:
\[ h = d_1 - \frac{d_1^2}{2(d_1 + d_2)} \]
Combining the terms gives:
\[ h = \frac{2d_1(d_1 + d_2) - d_1^2}{2(d_1 + d_2)} = \frac{d_1 d_2}{d_1 + d_2} \]

### Final Result
Thus, the height \( h \) of the wall in terms of \( d_1 \) and \( d_2 \) is:
\[ \boxed{h = \frac{d_1 d_2}{d_1 + d_2}} \]