A train moving with constant velocity $\frac{70}{9}{\mathrm{ms}}^{-1}$ when a bullet strikes it in a direction making angle ${37}^{\circ }$ with the train. The bullet enters one compartment at the corner farthest from the engine and passes out at the diagonally opposite corner. The compartment is 2.4 m long and 1.8 m wide. Choose the correct statement(s). 

(1) Let ABCD is the top view of compartment.

Here, AB = CD = 1.8 m 

and $\mathrm{BC}=\mathrm{AD}=2.4\mathrm{m}$

$\therefore \mathrm{AC}=\sqrt{(1.8{)}^2+(2.4{)}^2}=3\mathrm{m}$

and $\tan \mathrm{\alpha }=\frac{2.4}{1.8}=\frac{4}{3}$

$\Rightarrow \mathrm{\alpha }={53}^{\circ }$

Let v_b is actual velocity of bullet.

$\therefore {\mathrm{v}}_{\mathrm{bT}}={\mathrm{v}}_{\mathrm{b}}-{\mathrm{v}}_{\mathrm{T}}$

or ${\mathrm{v}}_{\mathrm{b}}={\mathrm{v}}_{\mathrm{bT}}+{\mathrm{v}}_{\mathrm{T}}$

$\Rightarrow {\mathrm{v}}_{\mathrm{b}}\cos {37}^{\circ }\hat{\mathrm{i}}+{\mathrm{v}}_{\mathrm{b}}\sin {37}^{\circ }\hat{\mathrm{j}}$

$={\mathrm{v}}_{\mathrm{bT}}\cos \mathrm{\alpha }\widehat{\mathbf{j}}+{\mathrm{v}}_{\mathrm{bT}}\sin \mathrm{\alpha }\widehat{\mathbf{j}}+\frac{70}{9}\widehat{\mathbf{i}}$

$\Rightarrow \frac{4}{5}{\mathrm{v}}_{\mathrm{b}}\widehat{\mathbf{i}}+\frac{3}{5}{\mathrm{v}}_{\mathrm{b}}\widehat{\mathbf{j}}=\frac{3}{5}{\mathrm{v}}_{\mathrm{bT}}\widehat{\mathbf{i}}+\frac{4}{5}{\mathrm{v}}_{\mathrm{bT}}\widehat{\mathbf{j}}+\frac{70}{9}\widehat{\mathbf{i}}$

$\therefore \frac{4}{5}{\mathrm{v}}_{\mathrm{b}}=\frac{3}{5}{\mathrm{v}}_{\mathrm{bT}}+\frac{70}{9}$

and $\frac{3}{5}{\mathrm{v}}_{\mathrm{b}}=\frac{4}{5}{\mathrm{v}}_{\mathrm{bT}}$

or $\frac{3}{4}{\mathrm{v}}_{\mathrm{b}}={\mathrm{v}}_{\mathrm{bT}}$

$\therefore \frac{4}{5}{\mathrm{v}}_{\mathrm{b}}=\frac{3}{5}\times \frac{3}{4}{\mathrm{v}}_{\mathrm{b}}+\frac{70}{9}$

$\Rightarrow \frac{4}{5}{\mathrm{v}}_{\mathrm{b}}-\frac{9}{20}{\mathrm{v}}_{\mathrm{b}}=\frac{70}{9}$

or $\left(\frac{16-9}{20}\right){\mathrm{v}}_{\mathrm{b}}=\frac{70}{9}$

$\Rightarrow {\mathrm{v}}_{\mathrm{b}}=\frac{70}{9}\times \frac{20}{7}=\frac{200}{9}{\mathrm{ms}}^{-1}$

$=\frac{200}{9}\times \frac{18}{5}=80{\mathrm{kmh}}^{-1}$

(3) $\therefore {\mathrm{v}}_{\mathrm{bT}}=\frac{3}{4}{\mathrm{v}}_{\mathrm{b}}=\frac{3}{4}\times 80=60{\mathrm{kmh}}^{-1}$

$\therefore \mathrm{t}=\frac{\mathrm{AC}}{{\mathrm{V}}_{\mathrm{bT}}}=\frac{3\times {10}^{-5}}{60}\mathrm{h}$

$\frac{3}{60}\times 60\times 60\mathrm{s}=180\mathrm{s}=3\min$