If a pitcher throws 65 mph from 54 ft. what would be the speed from 35.?

Question

I'm wondering the calculations.

Answer 1

What you mean to ask, I believe, is what would the speed be when the catcher catches a ball thrown at 65 mph from the two mounds. Clearly the ball thrown at 65 mph is the same no matter where the pitcher throws it.

The calculations you are looking for a very complex. What is clear, though, is that drag friction will act longer on the ball from the 54 ft. mound than from the 35 ft. one. This means the horizontal component (vx) of the 65 mph will be slowed down more from the 54 ft. mound because drag acts on the ball longer.

In fact, it is clear that the deceleration due to drag will be exactly the same for the first 35 feet of travel from the respective mounds. So we are talking about the additional deceleration of the x component over the extra 19 ft. the 54 ft. mound is from the plate. Thus, we can conclude the x component of the 65 mph velocity on toss will be greater for the 35 ft. mound than for the 54 ft. mound.

On the other hand, the vertical component (vy) of the 65 mph toss will be acted on by the force of gravity longer; so that vy = gt will be greater for the toss from 54 ft. than from 35 ft. In fact, as it was for the x component, the difference between the two mounds is that the ball will drop and accelerate longer from 54 feet than from 35 feet.

Velocity at any time t can be written as V(t) = sqrt(vy^2 + vx^2); so the question is this. Will the extra slowing time of the vx component be compensated for by the extra falling and accelerating time of the vy component for the 54 ft. mound over the 35 ft. mound?

Clearly, at t = 0, we have V(0) = 65 mph = sqrt(vy^2 + vx^2); so to stay exactly at 65 mph throughout a toss, the decrease in vx^2 must equal the increase in vy^2. Unfortunately, the calculations to determine these values over the full flight of the ball require complex calculus as the velocities and accelerations involved are time dependent.

I can tell you this, though. A Little League pitch from a mound about 2/3 the distance of a major league mound from home plate is equivalent in reaction time (distance/velocity = t) when the Little League pitch is multiplied by 1.3. For example, a Little League pitch of 60 mph = 1.3 X 60 = 78 mph for a major league pitch.

Answer 2

No matter the distance he is still throwing the same speed. Just the time to the plate is faster.