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  • A particle is ejected at a speed of 10  at an angle of 37 degrees from the horizontal. The radius of curvature at the time of projection is:Option

A particle is ejected at a speed of 10  at an angle of 37 degrees from the horizontal. The radius of curvature at the time of projection is:

Option: 1

12.78\ m


Option: 2

20.15\ m


Option: 3

50.6\ m


Option: 4

25\ m


Answers (1)

best_answer

Answer: (A) 12.78\ m

Explanation:

Radius of Curvature =  \frac{\left(\left\{1+\frac{dx}{dy}\right\}^2\right)^\frac{3}{2}}{\frac{d^2y}{dx^2}}

Equation for projectile is given as:

y=xtan\theta\ -\frac{1}{2}\frac{gx^2}{u^2\cos^2\theta}    

Therefore, the first and second derivatives are:

\frac{dx}{dy}=\ tan\theta\ -\ \frac{gx}{u^2\cos^2\theta}

And,

\frac{d^2y}{dx^2}=-\frac{g}{u^2\cos^2\theta}                     

Now,


\frac{\left(1+\tan^2\theta\right)^\frac{3}{2}\times u^2\times\cos^2\theta}{g}

Or,  \frac{u^2}{gcos\theta}              

 

Or,   \frac{10\times 10}{9.8\times \cos37}=\ 12.78              
 

Posted by

avinash.dongre

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