Two bodies are thrown up at angles of 45^o&nb | vedclass

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Question

(c) $H_{max} = \frac{{{u^2}{{\sin }^2}	heta }}{{2g}}$

According to problem $\,\,\,\frac{{{u_1}^2{{\sin }^2}45^\circ }}{{2g}} = \frac{{{u_2}^2{{\si

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$\sqrt {\frac{3}{2}} $

Vedclass · Answer

(c) $H_{max} = \frac{{{u^2}{{\sin }^2}	heta }}{{2g}}$

According to problem $\,\,\,\frac{{{u_1}^2{{\sin }^2}45^\circ }}{{2g}} = \frac{{{u_2}^2{{\sin }^2}60^\circ }}{{2g}}$

$ \Rightarrow \,\,\frac{{{u_1}^2}}{{{u_2}^2}} = \frac{{{{\sin }^2}60^\circ }}{{{{\sin }^2}45^\circ }}\,$ $ \Rightarrow \,\,\frac{{{u_1}}}{{{u_2}}} = \frac{{\sqrt 3 /2}}{{1/\sqrt 2 }} = \sqrt {\frac{3}{2}.} $

Two bodies are thrown up at angles of $45^o $ and $60^o $, respectively, with the horizontal. If both bodies attain same vertical height, then the ratio of velocities with which these are thrown is

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