A small object is projected horizontally from the top of a tower. If after 1 sec, The particle is at a height of 75 m from the ground, find its time of flight.

After 1 sec, vertical displacement of the particle 

$=\frac{1}{2}g{t}^2=\frac{1}{2}\times 10\times {\left(1\right)}^{2}m=5m$

$\therefore H=$Height of tower $=(5+75)m=80m$

If T be the time of flight, then $H=\frac{1}{2}g{T}^2\Rightarrow T=\sqrt{\frac{2\times 80}{10}}s=4\sec$