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

Question

The equation of motion of a projectile is $y=a x-b x^2$, where $a$ and $b$ are constants of motion. Match the quantities in Column I with the relations in Column II.

Professor PC Thomas Classes · Answer

(c) : Comparing given equation, \ $y=a x-b x^2$ with the equation of projectile motion \  $y=x 	an 	heta-\frac{g x^2}{2 u^2 \cos ^2 	heta}$, we get, $	an 	heta=a$...(i) $$ \begin{aligned} & 	ext { and } \frac{g}{2 u^2 \cos ^2 	heta}=b ...(ii) \ & 	herefore \quad \frac{g \sec ^2 	heta}{2 u^2}=b 	ext { or } \frac{g\left(1+	an ^2 	hetaight)}{2 u^2}=b \ & 	ext { or } u^2=\frac{g\left(1+	an ^2 	hetaight)}{2 b}  =\frac{g\left(1+a^2ight)}{2 b} \end{aligned} $$ or $\quad u=\sqrt{\left[\frac{g\left(1+a^2ight)}{2 b}ight]} \quad 	herefore \mathrm{A}-\mathrm{S}$ \ (B) Horizontal range, \ $R=\frac{u^2 \sin 2 	heta}{g}=\frac{2 u^2 \sin 	heta \cos 	heta}{g}$ $\begin{aligned} & R=\frac{2 u^2 \cos ^2 	heta}{g} 	imes 	an 	heta=\frac{a}{b} \ & 	herefore \quad \mathrm{~B}-\mathrm{P}\end{aligned}$ (Using (i) and (ii)) (C) Maximum height, $H=\frac{u^2 \sin ^2 	heta}{2 g}$ $H=\frac{u^2 \cos ^2 	heta}{2 g} 	imes 	an ^2 	heta=\frac{2 u^2 \cos ^2 	heta}{4 g} 	imes 	an ^2 	heta=\frac{a^2}{4 b}$ $ 	herefore \quad \mathrm{C}-\mathrm{R}  (Using (i) and (ii)) $ (D) From eqn. (ii), $u^2 \cos ^2 	heta=\frac{g}{2 b} \quad$ \  or $\quad u \cos 	heta=\sqrt{\frac{g}{2 b}}$ ...(iii) Time of flight $=\frac{2 u \sin 	heta}{g}=\frac{2 u \cos 	heta}{g} 	imes 	an 	heta$ $\begin{aligned} & =\frac{2}{g} \sqrt{\frac{g}{2 b}} 	imes a=a \sqrt{\frac{2}{b g}}  (Using (i) and (iii)) \ 	herefore \quad & \mathrm{D}-\mathrm{Q}\end{aligned}$