Still, a rock cannot be thrown forward from a tire that is encased by fenders. Look at a front tire and measure where a rock could possibly let go and hit the ground. There is a very small window of opportunity there. By the time the rock bounced up, the vehicle would be well past the rock.
This is, in my opinion, the most salient point in the whole discussion. The earlier statement that any object thrown from a tire must be moving at the same speed as the vehicle is incorrect. The average velocity of any part of the tire is identical to the vehicle's velocity. The instantaneous velocity is decidedly not, ranging from zero velocity to a velocity exactly double that of the vehicle. I'll illustrate later.
Can a rock be hurled by a tire faster than the vehicle itself?
The answer is an indisputable yes. It is only necessary to make an assumption that the rock would be held by the tire's tread. Once the assumption holds, then the rock will have the same velocity as the tangential velocity of the tire. Consider four points in the tire's rotation.
(1) At the point where the tread contacts the ground, the rock (and the tire) have zero velocity
in any direction. Is this really true? Yes. The road has no forward velocity, so at the point tire contacts the road, it too must have zero velocity. Earlier, it was stated that the rock hurled by the tire must have the same velocity as the vehicle. This is demonstrably false because it ignores tangential velocity. Any rock stuck in the tire tread under the contact patch has zero velocity, regardless of how fast the vehicle is travelling.
(2) At the very top of the tire, the tangential velocity of the tire (and any rock in its tread) is precisely double that of the vehicle. If the car is moving 70 MPH, then the top of the tire is traveling forward at 140 MPH. Any rock stuck in the tread at that point will also have a forward velocity of 140 MPH.
This illustrates an important point: the tangential velocity of a tire on a vehicle moving at 70 MPH is always 70 MPH. However, because the tire is in rotary motion, this motion can be broken into its horizontal and vertical components. The horizontal component of that velocity *must* be added or subtracted to the vehicle's velocity to compute the *forward* velocity at a point on the tire.
On the bottom of the tire, 70 MPH vehicle velocity - 70 MPH tangential horizontal tire velocity opposite that of the vehicle = 0 MPH forward motion at the tire tread. Any rock dislodged at this point would have no velocity at all.
On the top of the tire, 70 MPH vehicle velocity + 70 MPH tangential horizontal tire velocity in the same direction as the vehicle = 140 MPH forward motion. Any rock dislodged at this exact point would move exactly double that of the vehicle.
(3) At the rear of the tire (the point closest to the rear of the vehicle), the velocity is exactly identical to that of the vehicle. This is because the tangential velocity of the tire is purely in the vertical plane, meaning horizontal velocity is exactly zero.
70 MPH vehicle motion + 0 MPH tangential horizontal tire velocity = 70 MPH total forward velocity. Any rock dislodged at this point would have forward motion identical to that of the vehicle. It would also have massive vertical velocity, causing it to rocket skyward. It would not, however, hit the car (assuming there were no fender to inhibit its motion). The vehicle's speed is maintained by the power of its engine. The rock, now devoid of any motive force and experiencing the wind resistance of 70 MPH forward velocity, would slow down very quickly and land well behind the vehicle.
(4) At the front of the tire, the effect is identical to that of the rear of the tire except that any rock dislodged at this point will slam into the ground with a vertical velocity of 70 MPH. It will still have a forward velocity of 70 MPH.
This is the nature of a tire on a vehicle. The tread (any rock stuck in it) goes from zero forward velocity at the contact patch to double the vehicle's velocity at the top of the tire, passing each time through 70 MPH at the front and rear of the tire (defined relative to the vehicle). 0-70-140-70-0. Over and over and over again. If the vehicle has a constant motion so that the tire rotates at a constant rate, then the average forward velocity is exactly that of the vehicle. But to say that any rock dislodged from a tires tread can only move as fast as the vehicle is incorrect because instantaneous velocity <> forward velocity.
There was earlier a discussion about throwing a baseball from the back of a truck. Is this possible? Of course. I can throw a baseball about 60 MPH. If I stand in the back of a truck going 70 MPH and uncork one of my eminently hittable fastballs, the ball will leave my hand with the following velocities:
60 MPH relative to my hand and the truck.
130 MPH relative to the ground.
Upon release from my hand, it will encouter more ferocious wind resistance than any baseball has ever experienced from the hand of Aroldis Chapman. Lacking zero force in the direction it was propelled, it will decelerate at a massive rate. If I throw it just right, the truck will overtake it and the ball will hit the truck. (Or the truck will hit the ball, I suppose.)
Now, back to the salient point quoted in this post: given the degree of enclosure of most vehicle tires, it seems to me to be very difficult for this to happen. There are a few vehicles where the tires stick out of the fender wells to a significant extent. I suppose it might be possible to have a freak occurrence that discharges a rock just the right point, flinging it forward such that it bounces up and is caught by the windshield. But to happen twice? Dunno about that one.