Question

Show that the path of a projectile is a parabola.

Answer 1

1. Consider a body projected with velocity initial velocity `vec"u"`, at an angle θ of projection from point O in the coordinate system of the XY-plane, as shown in the figure.
   
2. The initial velocity `vec"u"` can be resolved into two rectangular components:
   u_x = u cos θ (Horizontal component)
   u_y = u sin θ (Vertical component)
3. The horizontal component remains constant throughout the motion due to the absence of any force acting in that direction, while the vertical component changes according to,
   v_y = u_y + a_yt
   with a_y = −g and u_y = u sinθ
4. Thus, the components of velocity of the projectile at time t are given by,
   v_x = u_x = u cosθ
   v_y = u_y −gt = u sinθ −gt
5. Similarly, the components of displacements of the projectile in the horizontal and vertical directions at time t are given by,
   s_x = (u cosθ)t   ...(1)
   s_y = (u sinθ)t − `1/2 "gt"^2`   ...(2)
6. As the projectile starts from x = 0, we can use
   s_x = x and s_y = y
   Substituting s_x = x in equation (1),
   x = (u cos θ)t
   ∴ t = `"x"/("u" cos theta)`   ...(3)
   Substituting, s_y = y in equation (2),
   y = (u sin θ)t -`1/2 "gt"^2`   ...(4)
   Substituting equation (3) in equation (4), we have,
   `"y" = "u"sin theta("x"/("u" cos theta)) - 1/2("x"/("u" cos theta))^2`g
   ∴ y = x (tan θ) - `("g"/(2"u"^2 cos^2 theta))"x"^2`   ...(5)
   Equation (5) represents the path of the projectile.

7. If we put tan θ = A and g/2u²cos²θ = B then equation (5) can be written as y = Ax - Bx² where A and B are constants. This is the equation of a parabola. Hence, the path of the projectile is a parabola.