A particle is fixed from point with a speed of 10 ms It hits

A particle is fixed from point with a speed of 10 m/s. It hits the ground 1 m vertically below point A. What it it\'s maximum height (m) above point A? Assume g = 10 m/s^2

Solution

From the figure

Angle = tan^-1 ( 8 /6) = 53.13

Maximum height reachead by the projectile is

            H = u² sin² ( 53.13 )/ 2 g

                =( 10 m/s)² (0.64 )/ 2 (10 m/s² )

                = 3.2 m

                =