1. Vertical motion of the projectile

At any time $t$, the vertical height is

$y = u\sin\theta\, t - \frac{1}{2}gt^2$

Here, $\theta = 30^\circ$, so

$y = u\sin 30^\circ\, t - 5t^2$

Since $\sin 30^\circ = \frac{1}{2}$,

$y = \frac{u}{2}t - 5t^2$

2. Same height at $t=3\,\text{s}$ and $t=5\,\text{s}$

Given that the projectile is at the same height at these two times,

$\frac{u}{2}(3) - 5(3)^2 = \frac{u}{2}(5) - 5(5)^2$

$\frac{3u}{2} - 45 = \frac{5u}{2} - 125$

3. Solve for $u$

Bring like terms together:

$-45 + 125 = \frac{5u}{2} - \frac{3u}{2}$

$80 = u$

So, the speed of projection is

$\boxed{80\ \text{m s}^{-1}}$

4. Alternative quick concept check

For a projectile, two times corresponding to the same height are symmetric about the time to reach maximum height.

So,

$t_{\text{up}} = \frac{3+5}{2} = 4\,\text{s}$

Now,

$t_{\text{up}} = \frac{u\sin 30^\circ}{g}$

$4 = \frac{u\cdot \frac{1}{2}}{10}$

$4 = \frac{u}{20} \Rightarrow u = 80\,\text{m s}^{-1}$

This confirms the result.