Physics, asked by manthanrajurkar17, 1 month ago

when a projectile is thrown at maximum range the equation of trajectory is​

Answers

Answered by sarahssynergy
0

when a projectile is thrown at maximum range the equation of trajectory is​

Explanation:

The trajectory of a projectile is given by -

                              y=xtanθ-12gx2u2cos2θ.

This equation is used for calculating diverse phenomenon inclusive of finding the minimum velocity required to make a stone reach a certain point at most range for a given projection velocity and the angle of projection required for maximum range.

Answered by brokendreams
0

For maximum range, the equation of trajectory is y = x - \frac{gx^{2} }{u^{2}}

Equation of path of the trajectory of a projectile

  • Consider a projectile motion of a body with velocity 'u' at an angle '\theta' from the horizontal axis against the gravitational field with the acceleration 'g'.
  • The two components of the velocity 'u' will be u_{x} = u \ cos \theta (horizontal component) and  u_{y} = u \ sin \theta (vertical component).
  • At the time 't' after projection,

x = u_{x} t \Rightarrow t = \frac{x}{u_{x}}  =  \frac{x}{u \ cos \theta} . . . . . . (1)

and, y = u_{y}t - \frac{1}{2} gt^{2} =  (u \ sin \theta) t - \frac{1}{2} gt^{2} . . . . . . (2)

  • Substituting (1) in (2), we get

y =  (u \ sin \theta) (\frac{x}{u \ cos \theta}) - \frac{1}{2} g[\frac{x}{u \ cos \theta}]^{2} \\\\\Rightarrow y = x \ tan \theta - \frac{g x^{2}}{2 u^{2} cos^{2}\theta } . . . . . . (3)

Calculating equation of trajectory at maximum range

  • At the maximum range, \theta = 45^{o}. Therefore, substituting \theta = 45^{o} in (3) to get

\Rightarrow y = x \ tan (45^{o} ) - \frac{g x^{2}}{2 u^{2} cos^{2} (45^{o} )  }\\\\\Rightarrow y = x \ (1) - \frac{g x^{2}}{2 u^{2} (\frac{1}{2})  }\\\\\Rightarrow y = x - \frac{g x^{2}}{u^{2}}\\

Hence, at maximum range, the equation of trajectory is​ y = x - \frac{gx^{2} }{u^{2}}

Similar questions