Suggested Answer
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.