Consider it being thrown with an angle #theta# with the horizontal.
Then the velocity has two components #u_x = Ucos theta# and #u_y = USin theta#.
The horizontal acceleration is #a_x = 0# while that in vertical direction is #a_y = -g#
From an equation from kinematics,
#y = u_yt + 1/2a_yt^2#
#implies y = USin thetat - 1/2 g t^2 # is distance travelled by particle in y direction (vertical direction) in time #t#.
Also, #x = u_xt + 1/2a_xt^2#
#implies x = UCos thetat#
Combining #x# and #y#,
#y = xtan theta - 1/2(gx^2)/Cos^2 theta#
But for #theta# constant, #tan theta = a# and #b = -g/(2Cos^2 theta)#,
We get,
#y = ax + bx^2#
Which shows that the path is parabolic.
The equations of motion are,
#F_x = ma_x = 0#
#F_y = ma_y = -mg#
If however, the body is thrown vertically up, #theta = pi/2#
And thats when the #x# component of motion ceases the exist and, motion can be described as, by acceleration #a_y = -g#.
Then, #y = Ut - 1/2 g t^2# describes the vertical distance travelled.
