A particle is projected with velocity 2 g h , so that it just flies over two walls of equal height h and 2 h distance apart from each other. Find the time for which the particle flies between the walls.

Step-by-step solution

Velocity of projection, aligned v & =2 g h v_x & =v =2 g h aligned Time taken by the projectile to cover interwall distance, aligned t & = 2 h v_x = 2 h 2 g h t & = h g aligned Vertical velocity at the top of the wall is given as aligned v_y^ 2 & =v_y^2-2 g h=( 2 g h )^2-2 g h & =4 g h ^2 -2 g h v_y^ 2 & =2 g h (2 ^2 -1 ) v_y^ & = 2 g h (2 ^2 -1 ) & t= 2 v_y^ g t & = 2 2 g h (2 ^2 -1 ) g aligned From Eqs. (i) and (ii), we get h g = 2 2 g h (2 ^2 -1 ) g Squaring both side, aligned & h g ^2 = 4 2 g h (2 ^2 -1 ) g^2 &