1. Resolve the initial velocity vertically

The particle is projected with speed $u=60\,\text{m/s}$ at angle $30^\circ$.

So the vertical component is $u_y=u\sin 30^\circ=60\cdot \frac{1}{2}=30\,\text{m/s}.$

2. Height traversed in the first second, $h_0$

Vertical displacement in time $t$ is $y=u_y t-\frac{1}{2}gt^2.$

For the first second, $t=1$ s: $h_0=30(1)-\frac{1}{2}(10)(1)^2=30-5=25\,\text{m}.$

3. Time to reach maximum height

At maximum height, vertical velocity becomes zero: $0=u_y-gt$ $0=30-10t$ $t=3\,\text{s}.$

So the particle reaches maximum height after $3$ seconds.

4. Height traversed in the last second before reaching maximum height, $h_1$

The last second before reaching the top is from $t=2$ s to $t=3$ s.

Compute heights at these times: $y(2)=30(2)-\frac{1}{2}(10)(2)^2=60-20=40\,\text{m}$ $y(3)=30(3)-\frac{1}{2}(10)(3)^2=90-45=45\,\text{m}$

Hence, $h_1=y(3)-y(2)=45-40=5\,\text{m}.$

5. Find the ratio

$h_0:h_1=25:5=5:1.$

Since the integer asked is the ratio value, $\frac{h_0}{h_1}=5.$

Therefore, the required answer is 5.