PROJECTILE MOTION

Consider a projectile from an initial position [x(0),y(0)].In this case acceleration (a) has only negative component ie, a_x=0 ; a_y= -g  .The components of displacement along the x axis and y axis are

x (t) = x(0) + v_x(0)t

y(t) = y(0) + v_y(0)t - ½ gt²

The path of the projectile is a parabola parallel to the projection of g. A particle projected with a velocity v(0) at an angle θ to the horizontal then horizontal range, maximum height, time of flight can be derived as follows

1. Maximum height, H:

For the motion in y direction v_y(0)=v(0)sinθ. At the maximum vertical height H,                                            

v_y(t) and y(t) - y(0) =H

        Using this eqn

v_y(t)²-v_y(0)² = 2 × -g × (y(t)-y(0))

0 - v(0)² sin ²θ   = -2g  H

H = v(0)²sin² θ / 2g

2. Time of flight, T:

It is time taken by the body to return to the same horizontal level as the point of projection. Then y(t) = y(0) and t = T.

From the eqn

y(t) = y(0) + v_y(0) t - ½ gt²

y(0) = y(0) + v(0)sin θT - ½ gT²

T = 2v(0) sin θ / g

3. Horizontal Range, R:

         Range on a plane is the distance between the point of projection and the point where the path of the projectile meets the horizontal plane through the point of projection.

      From the eqn

x(t) = x(0) + v_x(0)t

x(t) = x(0) + v(0)cos θT

Range, R = x(t) - x(0)

= v(0) cos θ×2×v(0) sin θ / g

= v(0)² sin2θ

To show that the path of the projectile is a parabola:

        Let v(0) be the initial velocity of projection and θ be the angle of projection. At time t = 0, let x(0) = 0 and y(0) = 0

v_x(0) = v(0) cos θ

v_x(0) = v(0) sin θ

The horizontal and vertical positions with respect to origin after time t is

x = v(0) cosθ × t

y =  (v(0) sin θ) t - ½ gt²

Substituting the value of t, from the eqn just above

y = xtanθ - gx² / 2×v(0)²cos²θ

This is the form y = bx - cx²  the eqn for the parabola. Hence the of the projectile is a parabola.