Question

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?

Answer 1

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.