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COLUMN-I (A) Angle of projection (B) Angle of velocity horizontal after 4 s (C) Maximum height (D) Horizontal range COLUMN-II (P) 20 m (Q) 80 m (R) 45^@ (S) tan^(-1)(0.5) |
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Answer» Given `y-x^(2)/(80)," comparing it with y"=x TAN THETA-(qx^(2))/(2u^(2) cos^(2) theta)" we have tan theta=1 rArr theta=45^(@)` We have `(2u^(2) cos^(2) theta)/(g)=80 rArr (2u^(2) (1//2))/(10)=80. ("where "theta " angle of projection & U: velocity of projection")` `rArr u^(2)=800 rArr u=20sqrt2m//s rArr u_(x)=u cos theta=20m//s and u_(y)=u sin theta =20m//s` `rArr" after 4 sec ", v_(y)=u_(y)+a_(y)t=20-40=-20m//s` `rArr` angle made by velocity vector with horizontal `=tan^(-1)""(v_(y))/(u_(x))=45^(@)`below horizontalMaximum height `=(u_(y)^(2))/(2g)=20m"Horizontal range "=(2u_(x)u_(y))/(g)=(2 XX 20 xx 20)/(10)=80` |
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