AP EAPCET 24-Aug-2021 Shift 2 Paper

Let $u$ and $\theta$ be the velocity of projection and angle of projection, respechively. Given that, horizontal component of velocity. $u_{x}=u \cos \theta=3\ \mathrm{m}/\mathrm{s}$ Equation of projectile motion is given by $y=12x-\frac{3}{4}x^{2}$ ...(i) We know, General equation of projectile motion, $y=x \tan \theta-\frac{g x^{2}}{2 u^{2} \cos^{2} \theta}$ ...(ii) Comparing Eqs. (i) and (ii), we get tanθ = 12 $\Rightarrow \frac{\sin \theta}{\cos \theta}=12$ $\sin \theta=12 \cos \theta$ Multiplying on both side by $(u)$ $u \sin \theta=12(u \cos \theta)=12 \times 3$ i.e. $u \sin \theta=36\ \mathrm{m}/\mathrm{s}$ Now, using the expression of range. $R=\frac{u^{2} \sin 2\theta}{g}$ $R=\frac{2 u^{2} \sin \theta \cos \theta}{g}$ $R=\frac{2(u \sin \theta)(u \cos \theta)}{g}$ [Using identity $\sin 2\theta=2 \sin \theta \cos \theta$ ] Substituting the values, we get $R=\frac{2 \times 36 \times 3}{10}$ = 21.6 m