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A projectile is projected with speed u at an angle of 60 degrees with the horizontal from the foot of an inclined plane.If the projectile hits the plane horizontally,the range on the inclined plane will be:
A)((u^2)*underroot 21)/2g
B)3u^2/4g
C)u^2/2g
D)underroot 21*u^2/8g
ANS:D
Dear Sameer,time = 1/2 . time of flight=usin(o)/g=u.sin60/gx axis distance = ucos60.t=ucos60.usin60/g=(u^2)sin60.cos60/gdistance on y=H(max)=(u^2)sin^2 (60)/2gHence total displacement=Range on inclined plane=root(x^2+y^2)=root(21)u^2/8g(ANS)
Regards.
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