Question 165: Motion in 2D & Vectors

Question

A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution. If this particle were projected with the same speed at an angle '$$	heta$$' to the horizontal, the maximum height attained by it equals 4R. The angle of projection, $$	heta$$, is then given by :

Accepted Answer

$$T = {{2\pi R} \over V}$$
$$V = {{2\pi R} \over T}$$
$$H = {{{u^2}{{\sin }^2}	heta } \over {2g}}$$
$$4R = {{4{\pi ^2}{R^2}{{\sin }^2}	heta } \over {{T^2}2g}}$$
$${\sin ^2}	heta  = {{8R{T^2}g} \over {4{\pi ^2}{R^2}}}$$
$${\sin ^2}	heta  = \sqrt {{{2{T^2}g} \over {{\pi ^2}R}}} $$
$$	heta  = {\sin ^{ - 1}}{\left( {{{2{T^2}g} \over {{\pi ^2}R}}} ight)^{1/2}}$$

Answer

$$	heta  = {\sin ^{ - 1}}{\left( {{{2g{T^2}} \over {{\pi ^2}R}}} ight)^{1/2}}$$

Answer

$$	heta  = {\cos ^{ - 1}}{\left( {{{g{T^2}} \over {{\pi ^2}R}}} ight)^{1/2}}$$

Answer

$$	heta  = {\cos ^{ - 1}}{\left( {{{{\pi ^2}R} \over {g{T^2}}}} ight)^{1/2}}$$

Answer

$$	heta  = {\sin ^{ - 1}}{\left( {{{{\pi ^2}R} \over {g{T^2}}}} ight)^{1/2}}$$

Motion in 2D & Vectors

A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution. If this particle were projected with the same speed at an angle '$\theta' to the horizontal, the maximum height attained by it equals 4R. The angle of projection, \theta$, is then given by :

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