The equation of motion of a projectile is y=a x-b x^2 , where a, b...

Matching the Columns i) The initial velocity of projection From our derivation, (u=$\sqrt{\frac{g\left(1+a^2\right)}{2 b}}$. This matches option d. ii) The horizontal range of a projectile The horizontal range, $R$, is the value of $x$ when $y=0$. $y=a x-b x²=x(a-b x)=0$ The solutions are $x=0$ and (x=$\frac{a}{b}$. The range is the non-zero solution. (R=$\frac{a}{b} .$ This matches option a. iii) The maximum height attained by the projectile The maximum height, $H$, occurs at the midpoint of the range, so at (x=R / 2=$\frac{a}{2 b}$. Substitute this into the trajectory equation: (H=a$\left(\frac{a}{2 b}\right)-$b\left(\frac{a}{2 b}\right)^$2=$\frac{a^2}{2 b}-$b\left(\frac{a^2}{4 b^2}\right)=\frac{a^2}{2 b}-\frac{a^2}{4 b}=\frac{2 a^2-a^2}{4 b}=\frac{a^2}{4 b} .This matches option c. iv) The$time of flight of the projectile The time of flight, $T$, is related to the range by the horizontal velocity: (R=u \cos$\theta \cdot$T). (T=$\frac{R}{u \cos \theta}=\frac{a / b}{u / \sqrt{1+\tan ^2 \theta}}=\frac{a}{b} \frac{\sqrt{1+\alpha^2}}{u} .$ Substitute our expression for $u$ : (T=$\frac{a}{b} \frac{\sqrt{1+a^2}}{\sqrt{\frac{\left(1+a^2\right)}{2}}}=\frac{a}{b} \sqrt{\frac{\left(1+a^2\right) 2 b}{g\left(1+a^2\right)}}=\frac{a}{b} \sqrt{\frac{2 b}{g}}=\frac{a \sqrt{2 b}}{b \sqrt{g}}=\frac{a}{b} \frac{\sqrt{2}}{\sqrt{b} \sqrt{g}}=\frac{a \sqrt{2}}{\sqrt{b} \sqrt{g}}=$a $\sqrt{\frac{2}{b g}} .$ This matches option b. Final Matches - i $→$ d - ii $→$ a - iii $→ c$ - iv $→$ b This matches the configuration of option (3).