question_answer

A body A of mass m is thrown with velocity \[\upsilon \] at an angle \[\mathbf{30}{}^\circ \] to the horizontal and another body B of the same mass is thrown with velocity \[\upsilon \] at an angle of \[\mathbf{60}{}^\circ \] to the horizontal, find the ratio of the horizontal range and maximum height of A and B.

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 \[{{H}_{A}}:{{H}_{B}}=\mathbf{1:3}\]