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?

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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: tv = u + at  =>  0 = u sin Ф - g t    =>  t = u sinФ / gtotal time it takes to reach back the ground :  2 t = 2 u sin Ф / grange of projectile :  speed * time  =  u cosФ * 2 u sin Ф/g  = u² sin 2Ф /gMaximum range for any direction:  when sin 2Ф = 1   => Ф = 45 deg.,                 maximum range =  u² /gSo time taken for projectile to go up & down: 2 u / g √2      as sin 45 = 1/√2         = √2 u /gdistance traveled vertically by a freely falling body in that time :            1/2  g  t²  =  1/2 g  2 u²/g²     = u²/gHence proved.