A projectile is fired from horizontal ground with speed v and projection angle θ . When the acceleration due to gravity is g , the range of the projectile is d . If at the highest point in its trajectory, the projectile enters a different region where the effective acceleration due to gravity is g ' = g 0 . 81 , then the new range is d ' = n d . The value of n is

Step-by-step solution

Horizontal range of the projectile is d = v 2 sin 2 θ g Maximum height attained by the projectile is H = v 2 sin 2 θ 2 g So, after entering the new region, time taken by projectile to reach ground t = 2 H g ' = 2 v 2 sin 2 θ 2 g g 0 . 81 = 0 . 94 v sin θ g So, horizontal displacement done by the projectile in new region is x = 0 . 9 v sin θ g × v cos θ = 0 . 9 v 2 sin 2 θ 2 g So, new range is d ' = d 2 + x = v 2 sin 2 θ 2 g + 0 . 9 v 2 sin 2 θ 2 g = 0 . 95 &