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Answer 1 · 2017-07-16T16:36:04

A projectile is projected from the origin with initial velocity $u$ making an angle $theta$ with the horizontal as shown in the figure below.

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Equation of the trajectory is given by the kinematic expression

$s=ut+1/2at^2$ .

As $x$ and $y$ components of the velocity of projection are orthogonal to each other these can be treated independently.

In the vertical direction.

Noting that the gravity is acting against the direction of motion
we have

$y = (u sin ⁡ theta)t - 1/ 2 g t^2$ ......(1)

In the horizontal direction.
Ignoring air resistance

$x = (u cos ⁡ θ)t$ ........... (2)

To combine both equations.

Eliminate $t$ by substituting value of $t$ from (2) in (1). We get

$y = u sin ⁡ θ x/ (u cos ⁡ θ )− g /2 ( x /(u cos ⁡ θ) )^ 2$
$=>y = x tan ⁡ θ − (g x ^2) /(2 u^ 2 cos^ 2 ⁡ θ)$

Writing second term on the RHS in terms of $tan theta$ using trigonometric identity we get

$y= x tan ⁡ θ − (g x ^2)/( 2 v ^2 )( 1 + tan ^2 ⁡ θ )$

Is the required equation of trajectory.