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 ' θ ' to the horizontal, the maximum height attained by it equals 4 R. The angle of projection, θ, is then given by :

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 ' θ ' to the horizontal, the maximum height attained by it equals 4 R. The angle of projection, θ, is then given by : (A) θ= cos -1(( gT 2/π2 R ))1 / 2 (B) θ= cos -1((π2 R/g T2))1 / 2 (C) θ= sin -1((π2 R/ g T 2))1 / 2 (D) θ= sin -1((2 gT 2/π2 R ))1 / 2

Tardigrade Admin · Accepted Answer

Velocity of the particle v=(2 π R/T) When projected at angle θ H= maximum height =(v2 sin 2 θ/2 g)=4 R =(4 π2 R2/T2) ( sin 2 θ/2 g)=4 R sin 2 θ=(2 g R T2/π2 R2) sin 2 θ=(2 g T2/π2 R) sin θ=((2 g T2/π2 R))1 / 2 θ= sin -1((2 g T2/π2 R))1 / 2

$θ = cos^{- 1} (\frac{g T ^{2}}{π ^{2} R})^{1/2}$

$θ = cos^{- 1} (\frac{π ^{2} R}{g T ^{2}})^{1/2}$

$θ = sin^{- 1} (\frac{π ^{2} R}{g T ^{2}})^{1/2}$

$θ = sin^{- 1} (\frac{2 g T ^{2}}{π ^{2} R})^{1/2}$

Q. 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 ' θ ' to the horizontal, the maximum height attained by it equals 4R. The angle of projection, θ, is then given by :

θ=cos−1(π2RgT2​)1/2

θ=cos−1(gT2π2R​)1/2

θ=sin−1(gT2π2R​)1/2

θ=sin−1(π2R2gT2​)1/2

Velocity of the particle v=T2πR​ When projected at angle θ H= maximum height =2gv2sin2θ​=4R =T24π2R2​2gsin2θ​=4R sin2θ=π2R22gRT2​ sin2θ=π2R2gT2​ sinθ=(π2R2gT2​)1/2 θ=sin−1(π2R2gT2​)1/2

$θ = cos^{- 1} (\frac{g T ^{2}}{π ^{2} R})^{1/2}$

$θ = cos^{- 1} (\frac{π ^{2} R}{g T ^{2}})^{1/2}$

$θ = sin^{- 1} (\frac{π ^{2} R}{g T ^{2}})^{1/2}$

$θ = sin^{- 1} (\frac{2 g T ^{2}}{π ^{2} R})^{1/2}$