A ball is projected with velocity u at right angle to the slope which inclined at an angle `alpha` with the horizontal. The distance `x` along the inclined plane that it will travell before again striking the slope is-

Options

A. `(2u^2)/(g)cosalpha`
B. `(2u^2)/(g)tanalpha`
C. `(2u^2)/(g)(tanalpha)/(cosalpha)`
D. `(2u^2)/(g)(tanalpha)/(sinalpha)`

Explanation

To solve the problem of a ball projected at right angles to a slope inclined at an angle \( \alpha \) with the horizontal, we will follow these steps:

### Step 1: Understand the Problem
The ball is projected vertically upward with an initial velocity \( u \) at an angle \( \alpha \) to the horizontal. We need to find the distance \( x \) along the inclined plane that the ball travels before it strikes the slope again.

### Step 2: Identify Forces and Components
When the ball is projected, the gravitational force \( g \) acts downward. We can resolve this gravitational force into two components relative to the inclined plane:
- Perpendicular to the slope: \( g \cos \alpha \)
- Parallel to the slope: \( g \sin \alpha \)

### Step 3: Calculate Time of Flight
The time \( t \) taken for the ball to reach its highest point (where its vertical velocity becomes zero) can be calculated using the equation:
\[ v = u + at \]
Here, \( v = 0 \) at the highest point, \( a = -g \) (acting downwards), and \( u \) is the initial vertical velocity. Thus,
\[ 0 = u - gt \implies t = \frac{u}{g} \]
Since the ball will take the same time to come down, the total time of flight \( T \) is:
\[ T = 2t = 2 \frac{u}{g} \]

### Step 4: Calculate the Distance Along the Incline
The distance \( x \) along the incline can be calculated using the kinematic equation for displacement:
\[ s = ut + \frac{1}{2} a t^2 \]
In this case, the initial velocity along the incline is \( 0 \) (since it is projected vertically), and the acceleration along the incline is \( g \sin \alpha \). Therefore,
\[ x = \frac{1}{2} g \sin \alpha \left(2 \frac{u}{g}\right)^2 \]
Substituting the value of \( T \):
\[ x = \frac{1}{2} g \sin \alpha \cdot \frac{4u^2}{g^2} \]
Simplifying this gives:
\[ x = \frac{2u^2 \sin \alpha}{g} \]

### Step 5: Final Expression
Thus, the distance \( x \) along the inclined plane that the ball will travel before striking the slope again is:
\[ x = \frac{2u^2 \sin \alpha}{g \cos \alpha} \]