A projectile is thrown into air with velocity $15 \mathrm{~m} / \mathrm{s}$ at an angle $30^{\circ}$ with the horizontal. After what time its direction of motion is perpendicular to its initial direction? (Assume, $g=10 \mathrm{~m} / \mathrm{s}^2$ )

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 $\mathbf{v}_1=u \cos \theta \hat{\mathbf{i}}+u \sin \theta \hat{\mathbf{j}}$ Since, in projectile motion the horizontal component of velocity remains same, while the vertical component varies, so its velocity after time $t$, $\begin{aligned} & \mathbf{v}_2=u_y \hat{\mathbf{i}}+v_y \hat{\mathbf{j}} \\ & \mathbf{v}_2=u \cos \theta \hat{\mathbf{i}}+(u \sin \theta-g t) \hat{\mathbf{j}}\end{aligned}$ $\left[\because v_y=u_y-g t=u \sin \theta-g t\right] \ldots$ As, velocities are perpendicular, so $\mathbf{v}_1 \cdot \mathbf{v}_2=0$ From Eqs (i) and (ii), we get $\begin{array}{ll}\Rightarrow & (u \cos \theta \dot{\mathbf{i}}+u \sin \theta \mathbf{j}) \cdot(u \cos \theta \mathbf{i}+(u \sin \theta-g t) \mathbf{j})=0 \\ \Rightarrow & u^2 \cos ^2 \theta+u \sin \theta(u \sin \theta-g t)=0 \\ \Rightarrow & u^2 \cos ^2 \theta+u^2 \sin ^2 \theta-u \sin \theta \times g t=0 \\ \Rightarrow & u^2\left(\cos ^2 \theta+\sin ^2 \theta\right)-u g t \sin \theta=0 \\ \Rightarrow & u^2=u g t \sin \theta \quad\left(\because \sin ^2 \theta+\cos ^2 \theta=1\right) \\ \Rightarrow & t=\frac{u}{g \sin \theta}\end{array}$ Here, $u=15 \mathrm{~m} / \mathrm{s}, g=10 \mathrm{~m} / \mathrm{s}^2$ and $\theta=30^{\circ}$ $\therefore \quad t=\frac{15}{10 \times \sin 30^{\circ}}=\frac{15 \times 2}{10}=3 \mathrm{~s}$