Consider it being thrown with an angle thetaθ with the horizontal.
Then the velocity has two components u_x = Ucos thetaux=Ucosθ and u_y = USin thetauy=Usinθ.
The horizontal acceleration is a_x = 0ax=0 while that in vertical direction is a_y = -gay=−g
From an equation from kinematics,
y = u_yt + 1/2a_yt^2y=uyt+12ayt2
implies y = USin thetat - 1/2 g t^2 ⇒y=Usinθt−12gt2 is distance travelled by particle in y direction (vertical direction) in time tt.
Also, x = u_xt + 1/2a_xt^2x=uxt+12axt2
implies x = UCos thetat⇒x=Ucosθt
Combining xx and yy,
y = xtan theta - 1/2(gx^2)/Cos^2 thetay=xtanθ−12gx2cos2θ
But for thetaθ constant, tan theta = atanθ=a and b = -g/(2Cos^2 theta)b=−g2cos2θ,
We get,
y = ax + bx^2y=ax+bx2
Which shows that the path is parabolic.
The equations of motion are,
F_x = ma_x = 0Fx=max=0
F_y = ma_y = -mgFy=may=−mg
If however, the body is thrown vertically up, theta = pi/2θ=π2
And thats when the xx component of motion ceases the exist and, motion can be described as, by acceleration a_y = -gay=−g.
Then, y = Ut - 1/2 g t^2y=Ut−12gt2 describes the vertical distance travelled.
