The equation of motion of a projectile is y=a x-b x^2 , where a, b are constants. Match the column - I with column - II array ll Column - I & Column - II i) The initial velocity of projection & a) a b ii) The horizontal range of projectile & b) a 2 bg iii) The maximum height attained by projectile & c) a ^2 4 ~b iv) The time of flight of projectile & d) g (1+a^2 ) 2 b array

Step-by-step solution

Matching the Columns i) The initial velocity of projection From our derivation, (u= g (1+a^2 ) 2 b ). This matches option d. ii) The horizontal range of a projectile The horizontal range, (R ), is the value of (x ) when (y=0 ). (y=a x-b x^2=x(a-b x)=0 ) The solutions are (x=0 ) and (x= a b ). The range is the non-zero solution. (R= a b . ) This matches option a. iii) The maximum height attained by the projectile The maximum height, (H ), occurs at the midpoint of the range, so at (x=R / 2= a 2 b ). Substitute this