A particle A is projected vertically upwards. | vedclass

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Question

$\frac{{v_1^2}}{{2g}}\,\, = \,\,\frac{{v_2^2\,\,{{\sin }^2}\,\,45}}{{2g}}\,\,\,\,	herefore \,\,\,\,v_1^2\,\, = \,\,\frac{{v_2^2}}{2}$

$\frac{{{K_1

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$1 : 2$

Vedclass · Answer

$\frac{{v_1^2}}{{2g}}\,\, = \,\,\frac{{v_2^2\,\,{{\sin }^2}\,\,45}}{{2g}}\,\,\,\,	herefore \,\,\,\,v_1^2\,\, = \,\,\frac{{v_2^2}}{2}$

$\frac{{{K_1}}}{{{K_2}}}\,\, = \,\,\frac{{\frac{1}{2}\,\,mv_1^2}}{{\frac{1}{2}\,\,mv_2^2}}\,\, = \,\,\frac{{v_1^2}}{{v_2^2}}\,\, = \,\,\frac{1}{2}$

A particle $A$ is projected vertically upwards. Another particle $B$ is projected at an angle of $45^{\circ}$. Both reach the same height. The ratio of the initial kinetic energy of $A$ to that of $B$ is

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