A particle is projected with a speed u at an angle `theta` with the horizontal. What will be the speed of the particle when direction of motion of particle is at an angle `alpha` with the horizontal?

Options

A. `u cos theta sec alpha`
B. `u sec theta cos alpha`
C. `u cos theta tan alpha`
D. `u cos theta cot alpha`

Explanation

To find the speed of a particle projected at an angle θ with the horizontal when its direction of motion is at an angle α with the horizontal, we can follow these steps:

### Step 1: Understand the Components of Initial Velocity
When the particle is projected with a speed \( u \) at an angle \( \theta \):
- The horizontal component of the initial velocity \( u_x = u \cos \theta \)
- The vertical component of the initial velocity \( u_y = u \sin \theta \)

### Step 2: Analyze the Motion
The horizontal motion is uniform because there is no horizontal acceleration. The vertical motion is influenced by gravity, which acts downward with an acceleration \( g \).

### Step 3: Determine the Velocity Components at Angle α
When the particle is at an angle \( \alpha \) with the horizontal, we can express the velocity components as:
- Horizontal component \( v_x = v \cos \alpha \)
- Vertical component \( v_y = v \sin \alpha \)

### Step 4: Use the Conservation of Energy
The total mechanical energy of the particle remains constant (ignoring air resistance). The kinetic energy at the initial point is given by:
\[ KE_{initial} = \frac{1}{2} u^2 \]
At the point where the particle makes an angle \( \alpha \) with the horizontal, the kinetic energy is:
\[ KE_{final} = \frac{1}{2} v^2 \]

### Step 5: Relate the Vertical Motion to the Initial Conditions
The vertical velocity at the angle \( \alpha \) can be expressed as:
\[ v_y = u \sin \theta - g t \]
where \( t \) is the time of flight until the particle reaches the angle \( \alpha \).

### Step 6: Apply the Pythagorean Theorem
The speed \( v \) of the particle can be found using the relationship between the components:
\[ v^2 = v_x^2 + v_y^2 \]
Substituting the expressions for \( v_x \) and \( v_y \):
\[ v^2 = (v \cos \alpha)^2 + (u \sin \theta - g t)^2 \]

### Step 7: Solve for Speed v
To find the speed \( v \) when the particle is at angle \( \alpha \):
\[ v = \sqrt{(u \cos \theta)^2 + (u \sin \theta - g t)^2} \]

### Final Expression
The final expression for the speed \( v \) of the particle when its direction of motion is at an angle \( \alpha \) with the horizontal is:
\[ v = \sqrt{(u \cos \theta)^2 + (u \sin \theta - g t)^2} \]