A body is projected with a speed 'uâ at an angle 'θ' with the horizontal. The radius of curvature of the trajectory when it makes an angle ((θ/2) ) with the horizontal is (g - acceleration due to gravity)

Question

A body is projected with a speed 'uâ at an angle 'θ' with the horizontal. The radius of curvature of the trajectory when it makes an angle ((θ/2) ) with the horizontal is (g - acceleration due to gravity) (A) (u2 cos2 θ sec3 ( θ/2)/√3g) (B) (u2 cos2 θ sec3 ( θ/2)/2 g) (C) ( 2 u2 cos2 θ sec3 ( θ/2)/g) (D) (u2 cos2 θ sec3 ( θ/2)/g)

Tardigrade Admin · Accepted Answer

Correct answer is (d) (u2 cos2 θ sec3 ( θ/2)/g)

Q. A body is projected with a speed 'u’ at an angle $^{'} θ^{'}$ with the horizontal. The radius of curvature of the trajectory when it makes an angle $(\frac{θ}{2})$ with the horizontal is
(g - acceleration due to gravity)

$\frac{u ^{2} c o s ^{2} θ s e c ^{3} ( θ /2 )}{3 g}$

$\frac{u ^{2} c o s ^{2} θ s e c ^{3} ( θ /2 )}{2 g}$

$\frac{2 u ^{2} c o s ^{2} θ s e c ^{3} ( θ /2 )}{g}$

$\frac{u ^{2} c o s ^{2} θ s e c ^{3} ( θ /2 )}{g}$

Correct answer is (d) $\frac{u ^{2} c o s ^{2} θ s e c ^{3} ( θ /2 )}{g}$

Q. A body is projected with a speed 'u’ at an angle ′θ′ with the horizontal. The radius of curvature of the trajectory when it makes an angle (2θ​) with the horizontal is (g - acceleration due to gravity)

3​gu2cos2θsec3(θ/2)​

2gu2cos2θsec3(θ/2)​

g2u2cos2θsec3(θ/2)​

gu2cos2θsec3(θ/2)​