Question

$A\:projectile\:is \: projected\: from \: ground \\with\: initial \: velocity\:(u_{0} i +v_{0} j) m/s. \\ If \: acceleration\: due\: to\: gravity\: is\: a = -gj,\\ then \:what\: will\: be\: the\: maximum\: vertical\\ displacement\: for \: maximum\: horizontal\\displacement?$

Answer 1

Hello mate !!

Thanks for asking this question !!

your answer is =>

As we know that, the maximum horizontal Displacement is called as Range of Projectile and maximum vertical displacement is called as Height of projectile.

so, by using the relation for maximum height of projectile ,

$=>\:\boxed{H\: =\: \frac{u^{2}}{2g}}$

here, velocity of projectile is given as =>

$=>u\: =\: u_{0}i\: +\: v_{0}j$

so, the resultant velocity will be ,

$=>\:u\: =\sqrt{u_{0}^{2}\:+v_{0}^{2}}$

substitute value of "u" in above equation ,

$=>\:H\: =\: \frac{\sqrt{u_{0}^{2}\:+v_{0}^{2}}^{2}}{2g}$

your final answer will be =>

$=>\:\boxed{\bf{H\:=\:\frac{u_{0}^{2}\:+\:v_{0}^{2}}{2g}}}$