A particle is projected from ground at an angle θ with horizontal with speed u . DERIVE the formula of radius of curvature of it's trajectory at a point of projection.
[hint:- R = u^2/(g cosθ)]
plz guys... help...!!..
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❇ Centripetal acceleration a = v²/ R
R ( Range ) = u² / g cos θ
Acceleration ( a) = g, ( Gravitational acceleration)
V ² = u² cos² θ
g = u² cos ² θ / R
RADIUS OF CURVATURE ( R) = u² cos ² θ / g.
❇ Centripetal acceleration a = v²/ R
R ( Range ) = u² / g cos θ
Acceleration ( a) = g, ( Gravitational acceleration)
V ² = u² cos² θ
g = u² cos ² θ / R
RADIUS OF CURVATURE ( R) = u² cos ² θ / g.
meenakshi997sa:
u have to derive the equation i gave in hint
I had given the formula of radius curvature...
As asked by u..
Sorry for misinterpretation
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