Assertion : In projectile motion if particle is projected with speed u, then speed of particle at height h would be `(sqrt (u^2 - 2gh)).`
Reason : If particle is projected with vertical component of velocity `u_y` . Then vertical component at the height h would be `+- (sqrt (u_(y)^2) - 2gh).`

Options

A. (a)If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.
B. (b)If both Assertion and Reason are true and the Reason is not the correct explanation of the Assertion.
C. (c) If Assertion is true, but the Reason is false.
D. (d) If Assertion is false, but the Reason is true.

Explanation

To solve the question, we need to analyze both the assertion and the reason provided regarding projectile motion.

### Step-by-Step Solution:

1. **Understanding the Assertion**:
- The assertion states that in projectile motion, if a particle is projected with an initial speed \( u \), then the speed of the particle at height \( h \) would be given by:
\[ v = \sqrt{u^2 - 2gh} \]
- This formula is derived from the conservation of mechanical energy, where the initial kinetic energy is converted into kinetic energy at height \( h \) and potential energy.

2. **Deriving the Speed at Height \( h \)**:
- The initial kinetic energy when the particle is projected is:
\[ KE_i = \frac{1}{2} mu^2 \]
- At height \( h \), the potential energy gained is:
\[ PE = mgh \]
- The kinetic energy at height \( h \) is:
\[ KE_f = \frac{1}{2} mv^2 \]
- By conservation of energy:
\[ KE_i = KE_f + PE \]
\[ \frac{1}{2} mu^2 = \frac{1}{2} mv^2 + mgh \]
- Canceling \( m \) from the equation (assuming \( m \neq 0 \)):
\[ \frac{1}{2} u^2 = \frac{1}{2} v^2 + gh \]
- Rearranging gives:
\[ v^2 = u^2 - 2gh \]
- Thus, taking the square root:
\[ v = \sqrt{u^2 - 2gh} \]

3. **Understanding the Reason**:
- The reason states that if the particle is projected with a vertical component of velocity \( u_y \), then the vertical component of velocity at height \( h \) would be:
\[ v_y = \pm \sqrt{u_y^2 - 2gh} \]
- This is also derived from the same principle of conservation of energy, focusing only on the vertical motion.

4. **Deriving the Vertical Component**:
- The vertical component of the initial velocity is \( u_y \).
- Using the same conservation of energy approach:
\[ \frac{1}{2} mu_y^2 = \frac{1}{2} mv_y^2 + mgh \]
- Canceling \( m \):
\[ \frac{1}{2} u_y^2 = \frac{1}{2} v_y^2 + gh \]
- Rearranging gives:
\[ v_y^2 = u_y^2 - 2gh \]
- Thus, taking the square root:
\[ v_y = \pm \sqrt{u_y^2 - 2gh} \]

5. **Conclusion**:
- Both the assertion and the reason are true.
- However, the reason does not provide a correct explanation for the assertion because the assertion refers to the total speed at height \( h \), while the reason refers specifically to the vertical component of velocity.

### Final Answer:
Both the assertion and reason are true, but the reason is not the correct explanation of the assertion.