Question

a projectile is projected from horizontally with velocity u making an angle 45° with the horizontal dir3ction find the distance of highest point of the projectile from its starting point

Answer 1

Hey dear here is your answer!!!!!!

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To find the maximum height ;

$h = u ^{2} \sin ^{2} theta \\ \: \: \: \div 2g$
(Refer the attachment )

⭐⭐ Hope it helps u dear...⭐⭐