A body when projected at an angle ' ' with the horizontal reaches a maximum height ' H '. The time of flight of the body will be ( g= acceleration due to gravity)

Step-by-step solution

Time of flight in terms of maximum height The maximum height H and time of flight T for a projectile with initial velocity u at angle are given by: H = u^2 ^2 2g T = 2u g From the height equation, u^2 ^2 = 2gH , so u = 2gH . Substituting into the time equation yields: T = 2 g 2gH = 2 2H g This corresponds to option C.