B

To analyze the situation, we consider the motion of a particle thrown vertically upward under the influence of gravity.
Step 1: Understanding the Motion
The particle is projected with an initial speed (denote it as $v_0$). The acceleration due to gravity is denoted as $g$. At the maximum height, the speed will be zero.
Step 2: Using Kinematic Equations
We can use the equation of motion to find the height ($h$) reached by the particle:
$$v^2 = v_0^2 - 2gh$$
At maximum height, $v = 0$, hence
$$0 = v_0^2 - 2gh$$
which implies
$$h = \frac{v_0^2}{2g}$$
Step 3: Analyzing Speed at a Given Height
Using the same equation at a height $h$, the speed ($v_h$) is given by:
$$v_h = \sqrt{v_0^2 - 2gh}$$
When the particle reaches zero speed at the maximum height, it will also have the same speed at the same height on the way down, confirming that this aspect is correct.
Step 4: Evaluating the Options
- Option A states that the speed at a certain time is $v_h$. This is correct, as speed at maximum height is zero.
- Option B suggests speed at distance $y$ is incorrect. This is also true, as speed at a given distance can be calculated with respect to the energy conservation principle. Depending on the trajectory, the speeds at the same height on the way up and down will be equal; specifically, for the two heights, the speed would be $\sqrt{v_0^2 - 2g\cdot y}$.
- Option C explains that at a certain time and position, the particle will be at the same height when thrown upward and downward. This is indeed correct.
- Option D mentions that velocities at two instances (on the way up and down) are the same, which holds true by the conservation of energy.
Conclusion: Therefore, the incorrect alternative is B.