A particle is projected at t=0 with velocity u at angle θ with the horizontal. Then the ratio of the tangential acceleration and the radius of curvature at the point of projection is :-

Question

A particle is projected at t=0 with velocity u at angle θ with the horizontal. Then the ratio of the tangential acceleration and the radius of curvature at the point of projection is :- (A) (g2 s i n 2 θ /u2) (B) (2 g2 s i n θ /u2) (C) (g2 s i n θ /u2) (D) (g2 s i n 2 θ /2 u2)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-ytwifecuvkva6jnb.jpg" /> R.O.C.=(u2/ gsin θ ) at=gcosθ so (at/ R . O . C .)=(g cos θ /u2)× g

Q. A particle is projected at $t = 0$ with velocity $u$ at angle $θ$ with the horizontal. Then the ratio of the tangential acceleration and the radius of curvature at the point of projection is :-

$\frac{g ^{2} s in 2 θ}{u ^{2}}$

$\frac{2 g ^{2} s in θ}{u ^{2}}$

$\frac{g ^{2} s in θ}{u ^{2}}$

$\frac{g ^{2} s in 2 θ}{2 u ^{2}}$

$R . O . C . = \frac{u ^{2}}{g s in θ}$
$a_{t} = g cos θ$
so $\frac{a _{t}}{R . O . C .} = \frac{g cos θ}{u ^{2}} \times g$

Q. A particle is projected at t=0 with velocity u at angle θ with the horizontal. Then the ratio of the tangential acceleration and the radius of curvature at the point of projection is :-

u2g2sin2θ​

u22g2sinθ​

u2g2sinθ​

2u2g2sin2θ​