Question:
a particle is projected with speed u at an angle to the horizontal find the radius of curvature at highest point of its trajectory
Answer:
Radius of curvature for parabola is given as:
(1 + y’^2)^3/2/|y’’|........................(1)
(1 + y’^2)^3/2/|y’’|........................(1)
Now the trajectory motion of the projectile motion i:
y = xtan(thetha) – gx^2/2u^2cos^2(thetha)
y = xtan(thetha) – gx^2/2u^2cos^2(thetha)
differentiate it both sides w.r.t x =>
dy/dx = tan(thetha) – gx/u^2cos^2(thetha)
SImilarly find double derivate and put both the eqns. in eqn. (1)
Put thetha ------> thetha/2
Put thetha ------> thetha/2
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