A projectile is thrown with velocity $u$ making angle $\theta$ with the vertical. It just crosses the top of two poles each of height h after 1 sec and 3 sec respectively. The maximum height of projectile is

The correct option is B 19.6 m

The projectile is at height h at times 1 sec and 3 sec, means the total time of projectile is 4 sec, since it takes the same 1 sec to reach ground after that 3 secs.

So it will be at Maximum height at time t = 2 sec

$H=\frac{1}{2}g{t}^2$
$H=\frac{1}{2}\times 9.8\times 2\times 2$
H = 19.6 m