If a particle is projected with speed 20 ms^-1 making angle ${37}^{\circ }$ with horizontal. At $\mathrm{t}=\frac{10}{3}\mathrm{s}$, 

(2) $\mathbf{v}=\mathbf{u}+\mathbf{a}\mathrm{t}\text{ or }\mathbf{v}=\left(20\cos {37}^{\circ }\hat{\mathrm{i}}+20\sin {37}^{\circ }\hat{\mathrm{j}}\right)-10\mathrm{t}\hat{\mathbf{j}}$

$=\left(20\times \frac{4}{5}\widehat{\mathbf{i}}+20\times \frac{3}{5}\widehat{\mathbf{j}}\right)-\frac{10\times 10}{3}\widehat{\mathbf{j}}$

$=16\widehat{\mathbf{i}}+12\widehat{\mathbf{j}}-\frac{100}{3}\widehat{\mathbf{j}}=16\widehat{\mathbf{i}}-\frac{64}{3}\widehat{\mathbf{j}}$

$\therefore \mathbf{u}\cdot \mathbf{v}=\left|\mathbf{u}\right|\left|\mathbf{v}\right|\cos \mathrm{\theta }$

$\Rightarrow (16\widehat{\mathbf{i}}+12\widehat{\mathbf{j}})\cdot \left(16\widehat{\mathbf{i}}-\frac{64}{3}\widehat{\mathbf{j}}\right)=\left|\mathbf{u}\right|\left|\mathbf{v}\right|\cos \mathrm{\theta }$

$\Rightarrow 256-256=\left|\mathbf{u}\right|\left|\mathbf{v}\right|\cos \mathrm{\theta }$

$\therefore \cos \mathrm{\theta }=0$

$\Rightarrow \mathrm{\theta }=\frac{\mathrm{\pi }}{2}$

(3) The magnitude of velocity is $\left|\mathrm{v}\right|=\sqrt{{16}^2+{\left(-\frac{64}{3}\right)}^2}=\frac{80}{3}{\mathrm{ms}}^{-1}$