Question

A projectile projected from ground, with velocity v making angle 60' with vertical, attains maximum height H. The time interval for which projectile remains in air is
a) √H/2g
b)√H/g
c)√8H/g
d)√4H/g​

Answer 1

H=u²sin²theta/2g

H= g²T²/4×2g. {T=2u sin theta/2g}

8H/g=T²

T=√8H/√g