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Physics 12 Online
OpenStudy (anonymous):

the angle of projection with horizontal of a projectile launched upwards from level ground is30 degree the ratio of horizontal range to maximum height is???

OpenStudy (anonymous):

we know R=u^2sintheta2theta/g Hmax=u^2/2g by taking ration of both u get R/Hmax=sin2theta/2 R/hamx=2sin30degree/2 R/hmax=1/2

OpenStudy (ujjwal):

R/H = 4U/V where R is horizontal range H is max height U =u cos30 is horizontal component of velocity V=u sin30 is vertical component of velocity I guess you can proceed further yourself

OpenStudy (anonymous):

how r/h=4u/v

OpenStudy (anonymous):

sorry now i get it

OpenStudy (ujjwal):

ok.. :)

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