Question

a projectile is fired with a velocity V making an angle theta with the horizontal show that its trajectory is parabolic ​

Answer 1

To show that the path of a projectile is parabolic

$\boxed{\setlength{\unitlength}{1 cm}\thicklines\begin{picture}(6.65,3.2) \put(0.3,1){\line(1,0){6}} \qbezier(0.3,1)(3,3)(6.3,1)\put(0.3,1){\vector(1,1){1}} \qbezier(0.5,1.2)(0.65,1,1)(0.6,1)\put(0.77,1.065){ \theta }\put(0.5,1.8){u}\put(3.15,1.5){\vector(0,1){0.5}}\put(3.15,1.5){\vector(0,-1){0.5}}\put(3.3,1.4){H}\put(4.3,0.7){\vector(-1,0){4}}\put(4.3,0.7){\vector(1,0){2}}\put(3.3,0.3){R}\end{picture}}$

Velocity of the projectile along the :

• X axis is u cos∅

• Y axis is u sin∅

Now,

$\sf \: velocity = \dfrac{displacement}{time} \\ \\ \longmapsto \: \sf \: u \: \cos( \alpha ) = \dfrac{x}{t} \\ \\ \longmapsto \: \sf \:t = \dfrac{x}{u \: \cos( \alpha ) } - - - - - - - - (1)$

Using the Relation,

$\sf \: y = ut \: + \frac{1}{2} {at}^{2} \\ \\ \longrightarrow \: \sf \: y = \cancel{u} \: \sin( \alpha ) \times \dfrac{x}{ \cancel{u} \: \cos( \alpha ) } - \dfrac{g}{2} \times \bigg( \frac{x}{u \: \cos( \alpha ) } \bigg) {}^{2} \\ \\ \longrightarrow \: \boxed{ \boxed{ \sf \: y = x \: \tan( \alpha ) - \dfrac{g {x}^{2} }{2 {u}^{2} cos {}^{2}( \alpha) } }}$

Equation of a parabola is y = ax - bx²

$\implies \: \sf \: a = \tan( \alpha ) \: and \: b = - \dfrac{g}{2 {u}^{2} {cos}^{2}( \alpha ) }$

Thus,the trajectory of the projectile is a parabola