A particle is projected from ground at an angle θ with horizontal with speed u. The radius of curvature of its trajectory at maximum height from ground is?
Options are
A) u^2 sin2θ/g B) u^2 cos^2θ/g
C) u^2 sin^2θ/g D) u^2 sin^2θ/2g
Answers
Answered by
4
radius of curvature of a curve y = f(x) is given by:
![If\ y=f(x),\ \ R=| \frac{[1+y'(x)]^{\frac{3}{2}}}{y''} |\\\\if\ y=f(t),\ \ x=g(t),\ \ R=| \frac{[(x'(t))^2+(y'(t))^2]^{\frac{3}{2}}}{x'(t)*y''(t)-y'(t)*x''(t)} |\\\\Projectile:\ y=x\ tan\theta-\frac{gx^2sec^2\theta}{u^2}\\\\ y'(x)=tan\theta-\frac{gxsec^2\theta}{u^2},\ \ y''(x)=-\frac{gsec^2\theta}{u^2}\\\\Simplify\ to\ get\ R=| \frac{(u^2-2gy)^{\frac{3}{2}}}{gu\ cos\theta} |=\frac{v^3}{g\ v_x}\\\\.\ \ \ where\ v(x,y,t)=instantaneous\ speed,\ \ v_x=u\ cos\theta\\\\](https://tex.z-dn.net/?f=If%5C+y%3Df%28x%29%2C%5C+%5C+R%3D%7C+%5Cfrac%7B%5B1%2By%27%28x%29%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%7By%27%27%7D+%7C%5C%5C%5C%5Cif%5C+y%3Df%28t%29%2C%5C+%5C+x%3Dg%28t%29%2C%5C+%5C+R%3D%7C+%5Cfrac%7B%5B%28x%27%28t%29%29%5E2%2B%28y%27%28t%29%29%5E2%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%7Bx%27%28t%29%2Ay%27%27%28t%29-y%27%28t%29%2Ax%27%27%28t%29%7D+%7C%5C%5C%5C%5CProjectile%3A%5C+y%3Dx%5C+tan%5Ctheta-%5Cfrac%7Bgx%5E2sec%5E2%5Ctheta%7D%7Bu%5E2%7D%5C%5C%5C%5C+y%27%28x%29%3Dtan%5Ctheta-%5Cfrac%7Bgxsec%5E2%5Ctheta%7D%7Bu%5E2%7D%2C%5C+%5C+y%27%27%28x%29%3D-%5Cfrac%7Bgsec%5E2%5Ctheta%7D%7Bu%5E2%7D%5C%5C%5C%5CSimplify%5C+to%5C+get%5C+R%3D%7C+%5Cfrac%7B%28u%5E2-2gy%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%7Bgu%5C+cos%5Ctheta%7D+%7C%3D%5Cfrac%7Bv%5E3%7D%7Bg%5C+v_x%7D%5C%5C%5C%5C.%5C+%5C+%5C+where%5C+v%28x%2Cy%2Ct%29%3Dinstantaneous%5C+speed%2C%5C+%5C+v_x%3Du%5C+cos%5Ctheta%5C%5C%5C%5C "If\ y=f(x),\ \ R=| \frac{[1+y'(x)]^{\frac{3}{2}}}{y''} |\\\\if\ y=f(t),\ \ x=g(t),\ \ R=| \frac{[(x'(t))^2+(y'(t))^2]^{\frac{3}{2}}}{x'(t)*y''(t)-y'(t)*x''(t)} |\\\\Projectile:\ y=x\ tan\theta-\frac{gx^2sec^2\theta}{u^2}\\\\ y'(x)=tan\theta-\frac{gxsec^2\theta}{u^2},\ \ y''(x)=-\frac{gsec^2\theta}{u^2}\\\\Simplify\ to\ get\ R=| \frac{(u^2-2gy)^{\frac{3}{2}}}{gu\ cos\theta} |=\frac{v^3}{g\ v_x}\\\\.\ \ \ where\ v(x,y,t)=instantaneous\ speed,\ \ v_x=u\ cos\theta\\\\")
At maximum height v = u cosθ = v_x.
Hence, R = u² Cos² θ / g
At maximum height v = u cosθ = v_x.
Hence, R = u² Cos² θ / g
atharvasdeshpande:
thanks
Similar questions