Question

The maximum range of a projectile is 2 times its actual range. What is the angle of projection for actual range?

Answer 1

let the actual range be

$r = \frac{ {u}^{2}sin2x }{g}$

and, we know the the max range

$r_{max} = \frac{ {u}^{2} }{g}$

Hence, according to given condition

$r_{max} = 2r \\ \frac{ {u}^{2} }{g} = 2( \frac{ {u}^{2}sin2x }{g} ) \\ 1 = 2sin2x \\ \frac{1}{2} = sin2x = sin {30}^{o} = sin {(2 \times 15)}^{o} \\ hence \\ x = {15}^{o}$

now, actual range of the projectile–

$r = \frac{ {u}^{2}sin2x }{g} = \frac{ {u}^{2}sin {(2 \times 15)}^{o} }{10} \\ r = \frac{ {u}^{2}sin {30}^{o} }{10} = \frac{ {u}^{2 } \times \frac{1}{2} }{10} = \frac{ {u}^{2} }{20}$

hence, the angle of projection and the actual range of the projectile are 15° and u²/20 units respectively.