Question

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

Answer 1

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.

Answer 2

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​ $y = x - \frac{gx^{2} }{u^{2}}$