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A particle is projected with a velocity u at an angle (theta) with the horizontal. Find the radius of curvature of the parabola traced out by the particle at the point where its velocity makes an angle (theta/2) with the horizontal.

Rahul , 9 Years ago
Grade 12
anser 3 Answers
Vikas TU

Last Activity: 9 Years ago

Radius of curvature for parabola is given as:

(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)
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
get the answer.
Vidu

Last Activity: 8 Years ago

When u draw the diagram you`ll notice that vcostheata/2=ucosthetav=ucostheta/costheta/2R=v^2/gcostheta/2Put value of v to get u^2cos^2theta/gcos^3theta
British singh

Last Activity: 7 Years ago

A projectile of mass m is fried with velocity v from the point p at an angle 45 with the horizon.the magnitude of change in momentum when it passes through the point q on the same horizontal line on which p lies is-
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