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What is the radius of curvature of the p...

What is the radius of curvature of the parabola traced out by the projectile.Projected with a speed `u=sqrt30` at angle `theta=60^(@)` with the horizontal at a point where the particle velocity makes an angle `theta//2` with the horizontal ?

Text Solution

Verified by Experts

The correct Answer is:
A
As velocity along horizontal remains constant
`U cos theta=V cos (theta/2)`
Radius of curvature `r=v^(2)/a_(T)=(v^(2)cos^(2)theta)/(g cos^(2)(theta//2)`
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What is the radius of curvature of the parabola traced out by the projectile in which a particle is projected with a speed u at an angle theta with the horizontal, at a point where the velocity of particle makes an angle theta//2 with the horizontal.

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Knowledge Check

  • A projectile is projected with a speed u at an angle theta with the horizontal. What is the speed when its direction of motion makes an angle theta//2 with the horizontal

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    `u cos theta`
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    `u (2 cos (theta)/(2) - sec (theta)/(2))`
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    `u sec theta cos alpha`
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    `u cos theta tan alpha`
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    `u cos theta cot alpha`

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