A body projected from the ground reaches a point ‘X’ in its path after 3 seconds and from there it reaches the ground after further 6 seconds. The vertical distance of the point ‘X’ from the ground is (acceleration due to gravity =10ms^–2)

According to the question, Total time of flight (t) = 2u​/g

where, u is the initial vertical velocity.

or, 9 = 2u/g​ or,  u = 9g/2​

u = 45m/s

Since, in covering the vertical distance, g becomes negative,

⇒h = ut−1/2​gt²

h = 45×(3)−1/2​×10×(3)²

h = 135−45

h = 90m