If we are throwing two objects directly to the ground you are right.
So from our kinematic equations:
$$V_f = V_i + at$$
I would ask your teacher. What happens to the $V_f$ if $V_i=0$? Then Follow it up with what would $V_f$ be if $V_i$ was very large?
The initial velocity DOES have an effect here.
HOWEVER: Make sure that you are not misinterpreting the question! If you are throwing two objects in something that is sort of parabolic arc (i.e. a projectile) and you threw them at the same time, the time it would take for each one to hit the ground would be THE SAME.
This is because at their peak the objects both only have one thing accelerating them (gravity) and because at their peak their V = 0! This is very un-intuitive and you should look into exploring it!
If you had a very squishy object, it will exert a small force on the ground, whereas if you have a very hard object there will be a large force. From this it's clear that asking for the force is ambiguous, we're going to need to introduce some other variable.
As the question suggests, one thing we can do is include a variable, lets call it $\Delta t$, which tells us the duration of the collision between object and ground. The nicest way to do this is to write down the force equation you had:
$F(t) = \frac{dp}{dt}$
I've included the time on the left hand side to remind us that the force will change as a function of time over the course of the collision.What we can do is integrate this equation from $t=0$, the time of contact, to $t=\Delta t$, the time at which the object comes to rest. Then
$\int_0^{\Delta t} F(t) dt = \int_0^{\Delta t} \frac{dp}{dt} dt = \Delta p$
We can multiply and divide by $\Delta t$ to see that
$\Delta p = \Delta t \left(\frac{1}{\Delta t} \int_0^{\Delta t} F(t) dt \right) = \Delta t \,\, F_{avg,t}$
Presumably this is how you found $F = M \sqrt{2g h} / t$, but I wanted to be clear what $F$ meant in that equation.
As an alternative which doesn't use the time, we can exploit the work energy theorem:
$F(x) = \frac{dW}{dx}$
Again, integrate both sides, this time from $x=0$ to $x=\Delta x$, the total distance over which the collision occurs. This time we find
$\Delta x F_{avg,x} = W$
By the work energy theorem $W = \Delta E$ where $\Delta E$ is the change in energy of the object, so
$F_{avg,x} = \frac{1}{2\Delta x} M v^2$
This gives us a way to write down an average force without reference to the time. The tradeoff is that now we have the distance over which the collision occurs, and we find the force averaged over position rather than over time.
Best Answer
When the ball makes contact with the ground, the ground exerts a very large (upward) force on the ball for a very short interval of time. This large force causes the ball velocity to change direction from downward to upward, and translates into a large upward acceleration of very short duration. So there is no inconsistency with either the laws of physics or the laws of mathematics.
If the ground is rigid, once the ball makes contact with the ground, the leading edge of the ball comes to a full stop, but the remainder of the ball is still moving downward. The ground exerts a force on the ball, and the ball begins to compress. A compression wave travels upward through the ball. The portion of the ball within the compression zone is not longer moving, but the part of the ball beyond the compression zone is still moving downward. Eventaully, the compression zone encompasses the entire ball, and the entire ball has come to a stop. Next, the compression begins to release. First the part of the ball at the top decompresses, and the velocity of this material is then upward. The decompression wave then travels downward until the ball is fully decompressed, and the entire ball is now traveling upwards. At this point, the ball loses contact with the ground. All these events take place within a tiny fraction of a second.
This description is qualitative, but it captures the essential mechanistic features of what is happening.