The motion of the object is described by the kinematic equations for constant acceleration, where the acceleration is $a = -g$. The initial velocity is $v_0$ and the initial position is $y_0 = 0$. The upward direction is positive.

The velocity $v(t)$ and displacement $y(t)$ at any time $t$ are given by:

The direction of the initial velocity is positive. The direction of velocity and displacement at time $t$ is determined by the sign of $v(t)$ and $y(t)$, respectively.

[Q1] Velocity and displacement in the same direction as initial velocity. This requires both $v(t) > 0$ and $y(t) > 0$.

1. Condition for velocity: $$v(t) > 0 \implies v_0 - gt > 0 \implies t < \frac{v_0}{g}$$
2. Condition for displacement: $$y(t) > 0 \implies v_0 t - \frac{1}{2}gt^2 > 0 \implies t\left(v_0 - \frac{1}{2}gt\right) > 0$$ Since $t > 0$, we must have $v_0 - \frac{1}{2}gt > 0$, which gives $t < \frac{2v_0}{g}$.

For both conditions to be true, the time must satisfy both $t < v_0/g$ and $t < 2v_0/g$. The more restrictive condition is $t < v_0/g$. Therefore, the interval is $0 < t < \frac{v_0}{g}$.

[Q2] Velocity opposite to initial velocity, displacement same as initial velocity. This requires $v(t) < 0$ and $y(t) > 0$.

1. Condition for velocity: $$v(t) < 0 \implies v_0 - gt < 0 \implies t > \frac{v_0}{g}$$
2. Condition for displacement: $$y(t) > 0 \implies t < \frac{2v_0}{g} \quad \text{(from Q1)}$$

Both conditions are met when $\frac{v_0}{g} < t < \frac{2v_0}{g}$.

[Q3] Velocity and displacement opposite to initial velocity. This requires both $v(t) < 0$ and $y(t) < 0$.

1. Condition for velocity: $$v(t) < 0 \implies t > \frac{v_0}{g} \quad \text{(from Q2)}$$
2. Condition for displacement: $$y(t) < 0 \implies v_0 t - \frac{1}{2}gt^2 < 0 \implies t\left(v_0 - \frac{1}{2}gt\right) < 0$$ Since $t > 0$, we must have $v_0 - \frac{1}{2}gt < 0$, which gives $t > \frac{2v_0}{g}$.

For both conditions to be true, the time must satisfy both $t > v_0/g$ and $t > 2v_0/g$. The more restrictive condition is $t > 2v_0/g$.