Question

Equation of oblique projectile can be written as:

Answer 1

Hint: In oblique projectile, horizontal plane of projectile motion is at some angle with the x- axis. As we get the equation of motion of the oblique projectile, we will simplify it in terms of x, R (range) and the angle of the projectile. First, we will multiply sin terms in the second term of the right-hand side of the equation, then simplify to make a formula of range in the denominator. This gives another form of equation of motion.

Complete answer:
A projectile is thrown with the velocity u at an angle $\theta$ with the x-axis. The velocity u can be resolved into two components $u cos \theta$ component along X-axis and $u sin \theta$ component along Y-axis.

$u_{x} = u cos \theta$ and $u_{y} = u sin \theta$
Equation of trajectory is:
$ y = x tan \theta - \dfrac{g x^{2} }{2 u^{2}cos^{2} \theta }$
Multiply by $sin \theta$ in the second term of the right-hand side.
$ y = x tan \theta - \dfrac{ x^{2} sin \theta }{\dfrac{2 sin \theta u^{2}cos^{2} \theta }{g} }$
$\implies y = x tan \theta - \dfrac{ x^{2} sin \theta }{ cos \theta \dfrac{ u^{2}sin 2 \theta }{g} }$
$\implies y = x tan \theta - \dfrac{x^{2} tan \theta}{R}$, R is the range of projectiles.
$\implies y = x tan \theta \left( 1 - \dfrac{x}{R} \right)$
Equation of oblique projectile can be written as
$ y = x tan \theta \left( 1 - \dfrac{x}{R} \right)$

Note:
When a particle is dropped in the air with speed, the only force performing on it during its air time is the acceleration due to gravity acting vertically downwards. There is no acceleration in the horizontal, which means that the particle's velocity in the horizontal direction lives constantly.