Match list I with list II for a projectile.
`{:("List-I", "List-II"),((A)"For two angles " theta and (90- theta) "with same magnitude of velocity of projection",(e)(PhatiPhati)/(g)),((b)"Equation of parabola of projectile" y=Px-Qx^(2), (f)"Maximum height" =25% (P^(2))/(Q)),((C)"Radius of curvature of path of projectile projected with a velocity"(Phati+Qhatj) ms^(-1)"at highest point",(g)"Range = Maximum height"),((d)"Angles of projection" theta =Tan^(-1)(4),(h)"Range is same"):}`

Answer

To solve the problem of matching List I with List II for a projectile, we will analyze each item in List I and find the corresponding item in List II based on the principles of projectile motion.

### Step-by-Step Solution:

1. **For two angles θ and (90 - θ) with the same magnitude of velocity of projection**:
- The range of a projectile launched at angles θ and (90 - θ) is the same. This is because the sine function has the property:
\[ \sin(2\theta) = \sin(180^\circ - 2\theta) \]
- Therefore, the correct match is:
- **A → H** (Range is the same)

2. **Equation of parabola of projectile**:
- The equation of the trajectory of a projectile is given by:
\[ y = Px - Qx^2 \]
- The coefficients P and Q relate to the initial velocity and the acceleration due to gravity. The match for this is:
- **B → F** (Maximum height = 25% (P²/Q))

3. **Radius of curvature of path of projectile projected with a velocity (P + Qj) m/s at highest point**:
- The radius of curvature at the highest point of the projectile's path can be calculated using:
\[ R = \frac{(P^2)}{g} \]
- This corresponds to:
- **C → E** (Radius of curvature = (P²)/(g))

4. **Angles of projection θ = tan⁻¹(4)**:
- If θ = tan⁻¹(4), then we can analyze the relationship between maximum height (H) and range (R). For a projectile, the maximum height and range have a specific relationship:
\[ H = \frac{R}{4} \]
- Therefore, the correct match is:
- **D → G** (Range = Maximum height)

### Final Matches:
- A → H
- B → F
- C → E
- D → G