Two particles A and B projected along different directions from the same point P on the ground with the same velocity of 70 m/s in the same vertical plane. They hit the ground at the same point Q such that PQ = 480 m, then (take, g = 9.8 m/s²)

$\begin{array}{l}\mathrm{R}=\frac{{\mathrm{u}}^2\sin 2\mathrm{\theta }}{\mathrm{g}}=480\mathrm{m}\\ \frac{{70}^2\sin 2\mathrm{\theta }}{9.8}=480\\ \Rightarrow \sin 2\mathrm{\theta }=\frac{480\times 9.8}{70\times 70}=\frac{24}{25}\\ \Rightarrow \cos 2\mathrm{\theta }=\frac{7}{25}\\ \Rightarrow 2{\cos }^2\mathrm{\theta }-1=\frac{7}{25}\\ \Rightarrow {\cos }^2\mathrm{\theta }=\frac{16}{25}\\ \Rightarrow \cos \mathrm{\theta }=\frac{4}{5}\\ \Rightarrow \mathrm{\theta }={37}^{\circ }\end{array}$

So, angles of projection will be 37° and 53°. 

$\begin{array}{l}\frac{{\mathrm{T}}_1}{{\mathrm{T}}_2}=\frac{\frac{2\mathrm{usin}{37}^{\circ }}{\mathrm{g}}}{\frac{2\mathrm{usin}{53}^{\circ }}{\mathrm{g}}}=\frac{3}{4}\\ \frac{{\mathrm{H}}_1}{{\mathrm{H}}_2}=\frac{9}{16}\\ \frac{{\mathrm{u}}_{min1}}{{\mathrm{u}}_{min2}}=\frac{\mathrm{ucos}{37}^{\circ }}{\mathrm{ucos}{53}^{\circ }}=\frac{4}{3}\left(\because \mathrm{velocity} \mathrm{at} \mathrm{highest} \mathrm{point} =\mathrm{ucos\theta }\right)\end{array}$

$\mathrm{\beta }=\frac{\mathrm{\theta }+\mathrm{\phi }}{2}=\frac{37+53}{2}=45$