Answer:

To solve this problem, we can use the following equations for projectile motion:

* **Time to reach the highest point:**

```

t = (2 * v_y) / g

```

* **Maximum height:**

```

h_max = (v_y^2) / 2g

```

* **Range:**

```

R = (v_x * t)

```

where:

* `v_y` is the initial vertical velocity

* `g` is the acceleration due to gravity (9.8 m/s^2)

* `t` is the time to reach the highest point or the time of flight

* `h_max` is the maximum height reached

* `R` is the range

We can first find the initial vertical velocity of the arrow by using the following equation:

```

v_y = v_i * sin(theta)

```

where:

* `v_i` is the initial velocity (28 m/s)

* `theta` is the angle of projection (30 degrees)

Substituting known values, we get:

```

v_y = 28 m/s * sin(30 degrees) = 14 m/s

```

Now we can find the time to reach the highest point by using the first equation:

```

t = (2 * v_y) / g = (2 * 14 m/s) / 9.8 m/s^2 = 3.13 s

```

The time of flight is the same as the time to reach the highest point, so the arrow will take 3.13 seconds to reach the target.

The maximum height reached by the arrow can be found by using the second equation:

```

h_max = (v_y^2) / 2g = (14 m/s)^2 / 2 * 9.8 m/s^2 = 4.4 m

```

Therefore, the arrow will reach a height of 4.4 meters before it hits the target.