Question

Derive formula for the time period of projectile motion.

Answer 1

Suppose the projectile be thrown with a velocity $u$ at an angle $\theta$ from the horizontal.

Let 't' be the time taken to reach the top-most point. At this point, the vertical component of velocity will become zero.
Using the first equation of motion along vertical direction,
$v_v=u_v+a_{v}t$
$\Rightarrow 0=u\sin \theta -gt$
$\Rightarrow t=\frac{u\sin \theta }{g}$
The same time will be taken to reach ground from the highest point.

So, time of flight, $T=\frac{2u\sin \theta }{g}$