Question

A particle is projected with a velocity v=ai+bj from ground find the radius of the curvature of the trajectory of the particle at its highest point​

Answer 1

Answer:a^/g

Explanation:

Since, tanθ0=ba,cosθ0=aa2+b2−−−−−−√' substituting v0=a2+b2−−−−−−√ and cosθ=aa2+b2−−−−−−√

We have, rP=(a2+b2^(3/2)ga)

At the highest position Q, the velocity of the particle is vQ=v0cosθ0. since, it moves horizantlly at the highest point Q,a→n=g(⊥v→)−→−−−−−. Hence, the radius of curvature at Q is

rQ=v2Qan=v20cos2θg,

where, v0cosθ=vx=a (given)

Then, rQ=a2g

Answer 2

Given:

A particle is projected with a velocity v=ai+bj from ground.

To find:

Radius of curvature of the particle at its highest point.

Calculation:

Radius of curvature at any point refers to the radius of the instantaneous circle that can be imagined at that point.

For radius of curvature, we need to consider the tangential velocity and the perpendicular acceleration.

Let radius of curvature be r.

Velocity at highest point is equal to the horizontal component i.e. x component vector.

$\therefore \: r = \dfrac{ {(v_{x})}^{2} }{g}$

$= > \: r = \dfrac{ {(a)}^{2} }{g}$

Refer to the attached diagram.

So , final answer is :

$\boxed{ \sf{ \: r = \dfrac{ {a}^{2} }{g} }}$