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Equation of Trajectory of a Projectile

The equation of the trajectory of a projectile shows that this trajectory path represents a parabola. This means that a projectile follows a parabolic path.

derivation of the equation of the projectile motion path
figure 1: Showing initial velocity V0 and its components along the X and Y axes for a projectile motion (Velocity components when time t =0)

Equation of Trajectory of a Projectile

Referring to Figure 1 above, we can calculate and get the trajectory path of a projectile as follows.
Equation of Trajectory of a Projectile: y = (tanθ) x – (1/2) g. x2/(V0 cosθ)2

As you see this can be rewritten in the form y = ax + bx2 where a and b are constants.
This is again an equation that represents Parabola.

Hence we can say that trajectory of a projectile will follow a curve that is represented as Parabola.

Deriving the Equation of the Trajectory of a Projectile

Referring to Figure 1 above, the initial velocity component along X-axis = V0x = V0 cosθ and the initial velocity component along Y-axis = V0y = V0sinθ.
(Air resistance is taken as negligible)

At time T = 0, there is no displacement along the X and Y axes.

At time T=t, (i.e., for any time instant t)
Displacement along X-axis = x= V0x.t = (V0 cosθ). t ………… (1) and
Displacement along Y-axis = y = (V0sinθ ).t  – (1/2) g t2 …………(2)

From equation 1 we get: t = x/(V0 cosθ) ………….. (3)

Replacing t in equation 2 with the expression of t from equation 3:

y = (V0sinθ ). x/(V0 cosθ)  – (1/2) g  [ x/(V0 cosθ)]2 
or, y = (tanθ) x –   (1/2) g . x2/(V0 cosθ)2………..(4)

In the above equation g, θ, and V0 are constant.

So rewriting equation 4:
y = ax + bx2 where a and b are constants. This is an equation representing a parabola.
So we can say that the trajectory or motion path of a projectile is a parabola.
So once thrown at an angle (excluding the right angle) with the horizontal, a projectile will follow a curved path named Parabola.

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