Question

Two objects are thrown up at angles of ${45}^{\circ }$ and ${60}^{\circ }$ respectively, with the horizontal. if both objects attain same vertical height, then the ratio of magnitude of velocities with which these are projected is

• A. (a) $\sqrt{\frac{5}{3}}$
• B. (b) $\sqrt{\frac{3}{5}}$
• C. (c) $\sqrt{\frac{2}{3}}$
• D. (d) $\sqrt{\frac{3}{2}}$

Answer 1

The correct option is D (d) $\sqrt{\frac{3}{2}}$

The correct option is D

Given,

$Angle{45}^0,{60}^0$

So,

$\frac{v_{1}si{n}^2{\alpha }_1}{2g}=\frac{v_2^{2}si{n}^2{\alpha }_2}{2g}$

So,

$(\frac{v_1}{v_2}{)}^2=\frac{\frac{si{n}^2{\alpha }_1}{2g}}{\frac{si{n}^2{\alpha }_2}{2g}}$

Where ${\alpha }_1={45}^0,{\alpha }_2={60}^0$

$(\frac{v_1}{v_2}{)}^2=(\frac{\frac{\sqrt{3}}{2}}{\frac{1}{\sqrt{2}}}{)}^2$

$=\frac{\frac{3}{4}}{\frac{1}{2}}=\frac{3}{2}$

So,

$\frac{v_1}{v_2}=\sqrt{\frac{3}{2}}$

Thus, the ratio of magnitude of velocities with which these are projected is $\frac{v_1}{v_2}=\sqrt{\frac{3}{2}}$