1for sum 1
we have
mv2/R=mg
but v= ucosθ
hence R=u2cos2θ/g
SUM 1:::
a particle is projected at a speed u at anangle thita{-}
wid the horizonatal
consider a small part near the highest position and take it to b a approx circular arc
Wht is d radius of this circle....????this radius is called the radius of curvature at this point
.
SUM 2
:::
wht is the radius of curvature of the parabola
traced
out by d projectile in d previoous problem
at a point where the particle
velocity
makes an angle [thita/2]
wid the horizontal
plz explain the second problem while solving...
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UP 0 DOWN 0 0 2
2 Answers
1for sum 2
let the velocity of object be v
then vcosθ/2=ucosθ(as horizontal component does not change)
hence we get v
and acceleration of particle perpendicular to velocity =gcosθ/2
so
v2/R=gcosθ/2
hence R=u2cos2θ/gcos3θ/2
