Given the equation of the projectile's path: \( Y = Px - Qx^2 \). Comparing this with the standard equation of a trajectory \( Y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta} \), we have: (A) \( P = \tan \theta \) (B) \( Q = \frac{g}{2 u^2 \cos^2 \theta} \) (a) Range: The range \( R \) is the value of \( x \) when \( Y = 0 \). \( Px - Qx^2 = 0 \) \( x(P - Qx) = 0 \) \( x = 0 \) or \( x = \frac{P}{Q} \) So, the range \( R = \frac{P}{Q} \). Thus, a-i. (b) Maximum height: The maximum height \( H \) occurs at \( x = \frac{R}{2} = \frac{P}{2Q} \). Substituting this into the equation for \( Y \): \( H = P \left( \frac{P}{2Q} \right) - Q \left( \frac{P}{2Q} \right)^2 \) \( H = \frac{P^2}{2Q} - \frac{P^2}{4Q} = \frac{P^2}{4Q} \) Thus, b-iii. (c) Time of flight: From \( Q = \frac{g}{2 u^2 \cos^2 \theta} \), we have \( u^2 \cos^2 \theta = \frac{g}{2Q} \). The time of flight \( T = \frac{2 u \sin \theta}{g} \). From \( P = \tan \theta \), we have \( \sin \theta = \frac{P}{\sqrt{1 + P^2}} \) and \( \cos \theta = \frac{1}{\sqrt{1 + P^2}} \). \( u^2 \cos^2 \theta = \frac{u^2}{1 + P^2} = \frac{g}{2Q} \) \( u^2 = \frac{g(1 + P^2)}{2Q} \) \( T = \frac{2 u \sin \theta}{g} = \frac{2 u P}{g \sqrt{1 + P^2}} = \frac{2 P}{g \sqrt{1 + P^2}} \sqrt{\frac{g(1 + P^2)}{2Q}} \) \( T = \frac{2 P}{g} \sqrt{\frac{g}{2Q}} = P \sqrt{\frac{2}{gQ}} \) Thus, c-iv. (d) Tangent of projection: From \( P = \tan \theta \), the tangent of projection is \( P \). Thus, d-ii. Therefore, the correct matching is: a-i, b-iii, c-iv, d-ii.
