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+clc
+clear
+//Input data
+m=0.04//Mass of stone in kg
+vi=25//Initial velocity in m/s
+vf=0//Final velocity in m/s
+yi=0//Initial height in m
+
+//Calculations
+Ui=(m*9.81*yi)//Initial potential energy in J
+Ki=(1/2)*m*vi^2//Initial kinetic energy in J
+Etotal=(Ui+Ki)//Total energy in J
+h=(Etotal/(m*9.8))//Maximum height in m
+//when the stone is at (2/3)h, total energy is again same
+v=sqrt((Etotal-(m*9.8*(2/3)*h))/((1/2)*m))//Velocity at (2/3) of its maximum height in m/s
+
+//Output
+printf('Maximum height it will reach is %3.1f m \n Potential energy at that height is %3.1f J \n velocity when it is at the two-third of its maximum height is %3.2f m/s',h,Etotal,v)