Question

A particle is projected with a speed u at an angle θ with the horizontal. Consider a small part of its path near the highest position and take it approximately to be a circular arc. What is the radius of curvature of the parabola traced out by the projectile in the previous problem at a point where the particle velocity makes an angle θ/2 with the horizontal?

Answer 1

Now, We know that in case of the Projectile motion,  Horizontal velocity remains the constant throughout the motion.

∴ Horizontal velocity at first = u cosθ

At Highest Point, Resultant velocity = u cosθ [Since, vertical will be zero.]

Now, If the velocity vector is making an angle θ/2 with the horizontal at some point in its trajectory, then

Horizontal component = vcosθ/2

∴  vcosθ/2 = ucosθ

⇒ v = ucosθ/cosθ/2

∴ mv²/r = mg

⇒ r = v²/g

⇒ r = u²cos²θ/gcos²θ/2

Hope it helps.

Answer 2

Explanation:

Dear Student,

Please find below the solution to the asked query:

Given that the initial velocity of the particle is U. The object is projected at an angle θ. Then the horizontal and vertical components of the initial velocity are,

Ux=U cos θ&Uy=U sin θ

Let after t time of projection, the particle is making an angle θ2/. Therefore,

V cos (θ2/)=U cos θ⇒V=U cos θcos (θ2/)∴V sin (θ2/)=(U cos θcos (θ2/))×sin (θ2/)⇒V=(U cos θcos (θ2/))2+((U cos θcos (θ2/))×sin (θ2/))2−−−−−−−−−−−−−−−−−−−−−−−⎷⇒V=U cos θcos (θ2/)1+sin2 (θ2/)−−−−−−−−−−−√

Therefore, the radius of curvature is,

g cos θ=V2R⇒R=V2g cos (θ2/) ⇒R=⎡⎣⎢⎢⎢U cos θcos (θ2/)1+sin2(θ2/)√⎤⎦⎥⎥⎥2g cos (θ2/)⇒R=U2 cos2 θ(1+sin2(θ2/))g cos3 (θ2/)

Hope this information will clear your doubts about the topic.