Question

A projectile is thrown with an initial velocity of v= ai^ + bj^ , if the range of projectile is double the maximum height reached by it then
a = 2b. or. b = a. or. b = 2a or. b = 4a.

Answer 1

$range =2 maximum \: height \\ = > \frac{ {u}^{2} \sin(2 \alpha ) }{g} =2 \frac{ {u}^{2} \sin( \alpha ) \sin( \alpha ) }{2g} \\ = > \frac{2 {u}^{2} \sin( \alpha ) \cos( \alpha ) }{g} =2 \frac{ {u}^{2} \sin( \alpha ) \sin( \alpha ) }{2g} \\ = > 2 \cos( \alpha ) = \frac{ \sin( \alpha ) } \\ = > \tan( \alpha ) = \\ = > but \: \tan( \alpha ) = \frac{vy}{vx} \\ = > given \: that\: vx = a \: vy = b \\ = > \tan( \alpha ) = \frac{b}{a} \\ = > \frac{b}{a} = \\ = > b = 2a$
here alpha is angle of projection

Answer 2

Answer:

The correct answer is b=2a