As shown in the figure there is a particle of mass `sqrt(3) kg`, is projected with speed `10 m//s` at an angle `30^(@)` with horizontal `("take "g=10 m//s^(2))` then match the following

`{:(,"Column I",,,"Column II"),((A),"Average velocity" ("in " m//s) "during half of the time of flight, is",,(P),1/2),((B),"The time (in sec after which the angle between velocity",,(Q),5/2_(o) sqrt(13)),(,"vector and initial velocity vector becomes " pi//2 is,,,),((C),"Horizonal range (in m), is",,(R),5sqrt(3)),((D),"Change in linear momentum (in N-s) when particle is at",,(S),"At an angle of" tan^(-1)(1/(2sqrt(3)))),(,"highest point, is",,,"from horizontal"),(,,,(T),2):}`

Answer

`(A) rarr (Q, S); (B) rarr (T); (C) rarr (R); (D) rarr (R)`

Explanation

For (A) : `v_(av)=sqrt((v_(avx))^(2)+(v_(avy))^(2))=sqrt((10 cos 30^(@))^(2)+((10 sin 30^(@)+0)/2)^(2))=sqrt(75+25/4)=5/2sqrt(13) m//s`
Angle with horizontal `theta=tan^(-1) (v_(avy)/v_(avx))=tan^(-1)((5//2)/(5sqrt(3)))=tan^(-1)(1/(2sqrt(3)))`
For (B) : By using `vec(v)=vec(u)+vec(a)t` we have `u/(g t)=sin 30^(@)rArr t=10/((10)(1//2))=2`
For (C) : Horizonatl range (R)`=(u^(2) sin 2theta)/(g)=(100xxsqrt(3)//2)/(10)=5sqrt(3) m`
For (D) : Change in linear momentum `=m u_(y)=sqrt(3)xx10 sin 30^(@)=5sqrt(3) N-s`