We study the movement of the stone on two axes of coordination Ox and Oy, with the origin on the ground.On the Ox axis the stone has a uniform rectilinear motion with velocity vₓ= v₀
On the axis Oy the stone has a uniformly accelerated motion with the acceleration g and without the initial velocity v₀y = 0 (free fall), so that the equation of motion is y=h-gt²/2.
The trajectory equation is obtained by eliminating the time from the two equations, thus obtaining t=x/v₀=>y=h-gx²/2v₀².The trajectory is a parabola because y is a function of the second degree of x.
The stone reaches the ground when y = 0 =>t=√(2h/g)=√(2*200/9.81)=√(400/9.81)=√40.77=6.38 seconds
The stone falls at a distance d=v₀*t=9*6.38=77.42 meters
