Physics, asked by Yuvraj36231, 10 months ago

Show that for a projection angle θ, where 0° < θ <90°, the path of the projectile is a parabola.

Answers

Answered by Anonymous
6

AnswEr :

We have to show that the Trajectory of a projectile is Parabolic

Let the initial velocity of the projectile be u and ang!e of projection be alpha .

Resolving the initial velocity vector along X and Y axes :

  • Along X Axis : u cos∅

  • Along Y axis : u sin∅

Acceleration along the axes :

  • Along X axis : 0

  • Along Y axis : - g

Displacement along x - axis :

 \sf \: x = u_xt  +  \dfrac{1}{2}  {at}^{2}  \\  \\  \longmapsto \:  \sf \: x = (u cos  \:  \theta )t \\   \\  \longmapsto \:  \sf \: t =  \frac{x}{u cos \:  \theta }

Displacement along y - axis :

 \sf \: y = u_yt  +  \dfrac{1}{2} a {t}^{2}  \\  \\  \longrightarrow \:  \sf \: y = usin \:  \theta  \times  \dfrac{x}{ucos \:  \theta }  -  \frac{g {x}^{2} }{2 {u}^{2} cos {}^{2} \theta  }  \\  \\  \longrightarrow \:  \sf \: y = x \: tan \:  \theta -  \dfrac{g {x}^{2} }{ {2u}^{2} {cos}^{2} \theta   }

Equation of a parabola is given as :

y = ax - bx²

Thus,

  •  \sf \: a = tan \:  \theta
  •  \sf \: b =  -  \dfrac{g}{2  {u}^{2}  {cos}^{2} \theta  }

Thus,the Path of a Projectile is Parabolic

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