Question

a projectile is projected from ground with initial velocity u = ui + vj if acceleration due to gravity g is along negative y direction then find maximum displacement in x Direction?
please answer the question fast.​

Answer 1

Answer:

$\frac{ 2{u} v}{g}$

Explanation:

He is asking about range you know formula of range

$r = \frac{2{u}^{2} \sin( \alpha ) \cos( \alpha ) } {g}$

here u was net velocity and alpha is angle of projection . Substitute values .

$\sqrt{ {u}^{2} + {v}^{2} } \: \: and \: \: \tan( \alpha ) = \frac{v}{u}$