Question

Derive
path of projectal is parabola​

Answer 1

Need to Do :-

• Derive that path of projectile is Parabola .

SolutioN :-

$\large\underline\red{\sf Breif \ Introduction :- }$

• When object is thrown obliquely with certain initial velocity and whose part is determined by the gravitational force is called projectile.
• For ex - When a batsman hits a ball in the air , it follows projectile motion .

Projectile at an Angle :-

• $\sf \red{ u \ sin\theta } \ is \ the \ vertical \ component.$
• $\sf \red{ u \ cos\theta } \ is \ the \ horizontal\ component.$

$\thicklines\setlength{\unitlength}{1 cm}\begin{picture}(12,8)\put(0.2,0){\vector(0,1){5}}\put(3,0){\vector(1,0){3}}\put(3,0){\line(-1,0){2.8}}\put(0,-0.3){ \sf O }\qbezier(0.2,0)(2.5,4)(5.5,0)\put(5.5,-0.3){ \sf A }\put(2.75,0){\line(0,1){2}}\put(2.75 , 2){\vector( 1 ,0 ){1}}\put(2.74,2){\circle*{0.4}}\put(5.1,0.6){\circle*{0.4}}\put(2.75,-0.3){ \sf M }\put(0.201,0){\vector(1,2){1}}\qbezier(0.8,0)(0.2, 1)(0.2,0.5)\put(.7,0.2){ \theta }\put(3,3){ \boxed{\sf @Rishabh\ Ranjan } }\end{picture}$

So , let us take that , Distance travelled in horizontal dirⁿ be $\sf S_x$ , intial Velocity be $\sf u_x$ . So that now ,

$\sf\to s_x = u_x \times t \\\\\sf\to t = \dfrac{s_x}{u_x}$

Now using equation of motion we have ,

$\sf:\implies s_y = u_y t - \dfrac{1}{2}gt^2 \\\\\sf:\implies s_y = u_y\bigg( \dfrac{s_x}{u_x}\bigg) - \dfrac{1}{2} g\bigg( \dfrac{s_x}{u_x}\bigg)^2 \\\\\sf:\implies s_y = (s_x ) tan\theta - \dfrac{g s_x^2}{2u^2 cos^2 \theta}$

Now taking , $\sf s_y = y \ \& \ s_x = x$ we have ,

$\sf:\implies\boxed{\pink{\sf{ y = x tan\theta - \bigg( \dfrac{g}{2u^2 cos^2\theta}\bigg) x^2 }}}$

The above equation is similar to Y = Ax - Bx² , which is the equation of Parabola. Where ,

$\sf \to \purple{ A = tan\theta} \\\\\sf\to \purple{B = \dfrac{g}{2u^2 cos^2\theta} }$

Therefore the trajectory of projectile is parabola .