Question

Derive the equation of the path of a projectile.

Answer 1

Hint
Before deriving the equation of the path of a projectile, let’s discuss projectile motion in brief. A motion where the body travels in an arc when thrown with a certain velocity at a given angle to the horizontal is known as projectile motion. The projectile motion can be said to be the sum of two different types of motion, wherein the vertical component of the motion is an accelerated motion but the horizontal component of the motion is a motion with uniform velocity. Let us proceed with the derivation of the path of the projectile.
Formula Used: $s=ut+\dfrac{1}{2}a{{t}^{2}}$ , $displacement=velocity\times time$

Complete step by step answer

A diagrammatic representation of projectile motion is given above.
The acceleration along the y-axis is given as $-g$ where the negative sign denotes direction.
The horizontal component of the initial velocity of the body is $v\cos \theta$ and the vertical component of the initial velocity is $v\sin \theta$
The displacement along the x-axis is given as $X$ and the displacement along the y-axis is given as $Y$
Since the acceleration is acting along the y-axis, we’ll apply the second equation of motion to find the displacement.
The second equation of motion is given as $s=ut+\dfrac{1}{2}a{{t}^{2}}$ where $s$ is the displacement, $u$ is the initial velocity and $a$ is the acceleration
Substituting the values in the above equation, we get
$\begin{aligned} & Y=v\sin \theta (t)+\dfrac{1}{2}(-g){{t}^{2}} \\\ & \Rightarrow Y=v\sin \theta t-\dfrac{g{{t}^{2}}}{2}-equation(1) \\\ \end{aligned}$
The motion along x-axis is a motion with constant velocity so displacement can be given as $displacement=velocity\times time$
Substituting the values, we get
$X=v\cos \theta \times t-equation(2)$
Now we have to eliminate the time variable from the two equations. This is done by substituting the value of time from the second equation into the first equation. The equation can now be given as
$\begin{aligned} & Y=v\sin \theta \left( \dfrac{X}{v\cos \theta } \right)-\dfrac{1}{2}g{{\left( \dfrac{X}{v\cos \theta } \right)}^{2}} \\\ & \Rightarrow Y=X\tan \theta -\dfrac{g{{X}^{2}}}{2{{v}^{2}}{{\cos }^{2}}\theta } \\\ \end{aligned}$
This is the required equation of projectile motion.

Note
The equation of projectile motion is the equation of its trajectory. So if we know the x-component of the position of an object, we can find the y-component of the position by using the equation of the projectile motion. All the quantities involved in projectile motion can be directly or indirectly obtained from the equation of the trajectory of projectile motion.