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COLUMN-I (A) Angle of projection (B)...

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)`

Text Solution

Verified by Experts

The correct Answer is:
A-r, B-q, C-p, D-q
[A-r] [B-r] [C-p] [D-q]
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 horizontal Maximum height `=(u_(y)^(2))/(2g)=20m" Horizontal range "=(2u_(x)u_(y))/(g)=(2 xx 20 xx 20)/(10)=80`
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