**Trajectory of Projectile Motion is a Parabola**

If you go to a stadium to enjoy cricket game then you will see that the cricket ball thrown from the boundary is seen to be ascending first and then comes back to the ground in a curved path. Similar phenomena are observed in case of motion of bullet fired upward from a gun, motion of arrows, bomb thrown from plane— in all cases same type of trajectories are observed.

This type of curved motion is called motion of projectile and the path is called trajectory. It is a parabola. This type of motion is two dimensional motion. If we ignore air-resistance motion of a projectile in due gravity only path of projectile is always parabolic. When the projectile reaches the maximum height, its velocity becomes minimum. Again, the motion of the projectile at maximum height is one dimensional. Maximum distance that a projectile travels horizontally is called the range of the projectile. Acceleration along horizontal, a_{x} = 0, and Vertical acceleration a_{y} = – g. At the point of ejection the coordinate of the origin is x = 0, y = 0. Suppose a projectile is thrown from point O vertically at an angle of θ with initial velocity v_{0} (Figure). Initial velocity of the projectile v_{0} can be resolved into two components. One component along OX and the other one along OY.

The two components are;

**v _{x0} = v_{0} cos **and

**v**

_{y0}= sin θWe can count time from the instant of ejection i.e., if distance travelled in time t = 0 is x, then according to the equation of motion, x = v_{x0} t + ½ a_{x} t^{2}

x = v_{0} cos θt + 0 [Horizontal motion v_{x0} = v_{0} cos θ]

so, t = x / [v_{0} cos θ] … … … (1)

Again, as a_{x} = – g, so after time t this velocity of the projectile in the vertical direction,

v_{y} = v_{y0} – gt = v_{0} sin θ – gt … … … (2)

If after time t, the projectile reaches to a height y then,

y = v_{yo}t – ½ gt^{2} = v_{0} sin θ t – ½ gt^{2} … … … (3)

Resultant velocity at time t, v = √(v_{x}^{2} + v_{y}^{2})

If the resultant velocity makes an angle α with the horizontal, tan α = v_{y}/v_{x}

Putting the value of t in equation (3) we get,

y = (v_{0} sin θ) [x/(v_{0} cos θ)] – ½ g [(x^{2} / v_{0}^{2}cos^{2}θ)]

so, y = (tan θ)x – [(g /2 v_{0}^{2}cos^{2}θ) x^{2}

**y = ax — bx ^{2} **

**This is the equation of a parabola.**

Here a = tan θ, b = g /2 v_{0}^{2}cos^{2}θ

These two quantities are constant in the path of ejection. So** trajectory of a projectile is a parabola.**