Trajectory of a particle in a projectile motion is given by `y=x-x^(2)/80` where x and y are in meters. Match the column-1 and column-2.
`{:(,"Column-I",,"Column-II"),("(A)","x coordinate at height of 15 m",,"(P) 20 m"),("(B)",underset("point of projection at x= 100 m")"vertical distnce of particle from",,"(Q) 80 m"),("(C)","Horizontal range",,"(R) 60 m"),(,,,"(S) 25 m"):}`

Answer

[(A)PR(B)S(C)Q]

Explanation

(A)`15=x-x^(2)/80`
`1200=80x-x^(2)`
`x^(2)-80x+1200=0`
`x^(2)-60x-20x+1200=0`
`x=60,20`
(B) `y=100-(100)^(2)/80=100-(1000)/82=25`
(C) `Hz range:x-x^(2)/80=0Rightarrow x(x-x/80)=0,x=(0,80)`