Answer:
For body \[\text{A }:\text{ }\theta \text{ }=\text{ 3}0{}^\circ \] Horizontal range, \[{{R}_{A}}=\frac{{{u}^{2}}{{\sin }^{2}}(2\times {{30}^{\circ }})}{g}=\frac{{{u}^{2}}}{g}\times \frac{\sqrt{3}}{2}.\] Maximum height, \[{{H}_{A}}=\frac{{{u}^{2}}{{\sin }^{2}}{{30}^{\circ }}}{2g}=\frac{{{u}^{2}}}{2g}\times \frac{1}{4}\] For body \[B:\theta ={{60}^{\circ }}\] \[\therefore \] \[{{R}_{B}}=\frac{{{u}^{2}}\sin (2\times {{60}^{\circ }})}{g}=\frac{{{u}^{2}}}{g}\times \frac{\sqrt{3}}{2}\] and \[{{H}_{B}}=\frac{{{u}^{2}}{{\sin }^{2}}{{60}^{\circ }}}{2g}=\frac{{{u}^{2}}}{2g}\times \frac{3}{4}\] Hence \[{{R}_{A}}:{{R}_{B}}=\mathbf{1:1}\] and \[{{H}_{A}}:{{H}_{B}}=\mathbf{1:3}\]
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