This problem involves one-dimensional motion under constant acceleration due to gravity. We will use the standard kinematic equations to solve for the final velocity of the object.

Let's establish a coordinate system where the upward direction is positive and the ground is at $y=0$. The object is dropped from an initial height $h = 150 \text{ m}$, so its initial position is $y_0 = h$. Its final position is $y = 0$. The acceleration is constant, $a = -g$. The time of flight is $t = 6 \text{ s}$.

We can relate the final velocity $v$ to the displacement, acceleration, and time using a kinematic equation that does not require the initial velocity $v_0$:

$$y - y_0 = v t - \frac{1}{2} a t^2$$

[Q1] Find the velocity of the object when it hits the ground.

We first derive an expression for the final velocity $v$ from the kinematic equation above.

$$v t = (y - y_0) + \frac{1}{2} a t^2$$ $$v = \frac{y - y_0}{t} + \frac{1}{2} a t$$

Now, we substitute the values from our coordinate system: $y=0$, $y_0=h$, and $a=-g$.

$$v = \frac{0 - h}{t} + \frac{1}{2}(-g)t$$ $$v = -\frac{h}{t} - \frac{gt}{2}$$

Finally, we substitute the given numerical values: $h = 150 \text{ m}$, $t = 6 \text{ s}$, and $g = 10 \text{ m/s}^2$.

$$v = -\frac{150 \text{ m}}{6 \text{ s}} - \frac{(10 \text{ m/s}^2)(6 \text{ s})}{2}$$ $$v = -25 \text{ m/s} - \frac{60}{2} \text{ m/s}$$ $$v = -25 \text{ m/s} - 30 \text{ m/s}$$ $$v = -55 \text{ m/s}$$

The negative sign indicates that the object's velocity is in the downward direction, as expected.