To find the maximum height of the projectile, we can use the equations of motion for projectile motion. Let's analyze the situation step by step.

Initial velocity components:
The initial velocity of the particle can be split into two components: one along the vertical direction and one along the horizontal direction.

Vertical component:
The velocity of the particle in the vertical direction is given by v_y = u sin(θ), where u is the initial velocity magnitude and θ is the angle with the vertical.

Horizontal component:
The velocity of the particle in the horizontal direction is given by v_x = u cos(θ).

Time of flight:
The time taken by the particle to cross the top of the first pole is 1 second, and to cross the top of the second pole is 3 seconds. The time of flight (T) can be calculated as the sum of the time taken to reach the maximum height (t_rise) and the time taken to fall from the maximum height (t_fall).

t_rise + t_fall = T
t_rise + t_fall = 1 + 3 = 4 seconds

Using the equation of motion for vertical motion, we can find the time taken to reach the maximum height (t_rise) as:

t_rise = v_y / g
t_rise = (u sin(θ)) / g

Similarly, the time taken to fall from the maximum height (t_fall) is also t_rise.

t_fall = t_rise = (u sin(θ)) / g

So, the total time of flight (T) is:

T = t_rise + t_fall
T = 2(t_rise)
T = 2[(u sin(θ)) / g]

Maximum height:
The maximum height (H) reached by the particle can be found using the equation of motion for vertical motion:

H = (v_y)^2 / (2g)
H = (u sin(θ))^2 / (2g)

Substituting the value of v_y = u sin(θ), we get:

H = (u sin(θ))^2 / (2g)
H = (u^2 sin^2(θ)) / (2g)

Since we know that T = 2[(u sin(θ)) / g], we can express sin^2(θ) in terms of T:

sin^2(θ) = (Tg / 2u)^2

Substituting this value in the equation for maximum height, we get:

H = (u^2 sin^2(θ)) / (2g)
H = (u^2 (Tg / 2u)^2) / (2g)
H = (T^2g^2) / (8u^2)

We know that the height of each pole is h, and the particle just crosses the top of each pole. Therefore, the maximum height reached by the particle is equal to twice the height of the pole:

H = 2h

Setting the equations for H equal to each other, we can solve for h:

2h = (T^2g^2) / (8u^2)
h = (T^2g^2) / (16u^2)

Since g = 9.8 m/s^2, T = 4 seconds, and u is the initial velocity magnitude,