TS EAMCET 9-Sep-2020 Shift 2 Solved Paper

The given projectile motion is as shown in figure, Let $t$ be the time after which the initial direction is perpendicular to its direction of motion. As, velocity of a projectile has two components So, its initial velocity is given by $v_{1}=u \cos \theta \hat{i}+u \sin \theta \hat{j}$.....(i) Since, in projectile motion the horizontal component of velocity remains same, while the vertical component varies, so its velocity after time $t_{\;\text{ , }\;}$ $v_{2}=u_{y}\hat{i}+v_{y}\overline{j}$ $v_{2}=u \cos \theta \hat{i}+(u \sin \theta-g t) \hat{j}$ $[\because v_{y}=u_{y}-g t=u \sin \theta-g t]$.....(ii) As, velocities are perpendicular, $50 v_{1} \cdot v_{2}=0$ From Eqs (i) and (ii), we get $\Rightarrow u^{2} \cos ^{2} \theta+u \sin \theta(u \sin \theta-g t)=0$ $\Rightarrow u^{2}(\cos ^{2} \theta+\sin ^{2} \theta)-u g t \sin \theta=0$ $\Rightarrow \;\; t=\frac{u}{g \sin \theta}$ Here, $u=15 \text{ m/s}, g=10 \text{ m/s}^{2}$ and $\theta=30^{\circ}$