Question 171: 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 :

Choose the correct answer: