Show that the max. range of a projectile in any direction is described in the same time in which it would fall freely under gravity through this distance starting from rest?

Let the projectile be launched with a speed u with an angle Ф.  Its vertical component is u sin Ф and horizontal component is u cos Ф.
Let the time it takes to reach the top height: t

v = u + at  =>  0 = u sin Ф - g t    =>  t = u sinФ / g
total time it takes to reach back the ground :  2 t = 2 u sin Ф / g

range of projectile :  speed * time  =  u cosФ * 2 u sin Ф/g  = u² sin 2Ф /g
Maximum range for any direction:  when sin 2Ф = 1   => Ф = 45 deg.,
                 maximum range =  u² /g

So time taken for projectile to go up & down: 2 u / g √2      as sin 45 = 1/√2
         = √2 u /g

distance traveled vertically by a freely falling body in that time :
           1/2  g  t²  =  1/2 g  2 u²/g²     = u²/g

Hence proved.

The correct answer is c