Hi. This is my first post. You guys have a great forum. Anyway, my question is about baseball. The pitcher's mound is approx. 60 feet from home plate. By the time the ball gets to home plate, it's usually traveling somewhere around 90 mph. My question is: What if homeplate was only 5 or 6 feet from the pitchers mound? Would the ball still be traveling at 90 mph when it reached home plate? Or, does it take a certain amount of distance to accelerate up to 90 mph? Thanks guys
The ball is traveling fastest when it leaves the pitcher's hand. It is slowed by air resistance as it moves toward home plate. I wonder what is meant when they announce that a tennis ball is served at 125 MPH or a baseball is pitched at 95 MPH Do they clock the time from pitcher to home plate and calculate the average speed? Do they clock a few feet of motion near home plate, giving them the speed the batter must contend with? Do they clock a few feet near the pitcher, giving them maximum speed as it leaves the pitcher’s hand? My guess is that they announce maximum speed as baseball leaves pitcher’s hand or ball leaves tennis racket. The media likes to quote the fastest, highest, et cetera. It sounds more impressive.
Something can only accelerate if theres a force acting on it. There is no force when flying in mid air so it slows down due to air resistance.
Dinosaur, Thanks for the reply. I think they use a radar gun similar to those that police use to nail speeders. When i watch baseball, sometimes the speed of the pitch will be posted even before it crosses home plate. I noticed that last night.
I don't have any numbers, but I would find it hard to believe that the speed changes much from pitch to bat.
Oh it would! Imagine an aircraft flying. It switches the engines off. Very soon the aircraft starts descending because it has slowed down so much it cannot provide enough airflow over the wings. Now remember that becasue of its mass, an aircraft has a lot of momentum so it would take a lot of force to slow it down. A baseball will be slowed down even more effectively.
OK, I have numbers now: regulations: mound to plate = 60'6" = 18.4 m 5 oz. < weight of ball < 5.25 oz. = 0.142 kg < m < 0.149 kg 9" < circumference of ball < 9.25" = 0.00416 m<sup>2</sup> < A < 0.00439 m<sup>2</sup> estimations: 60 mph = 27.0 m/s C = 0.5 (for spherical objects) ρ = 1.25 kg/m<sup>3</sup> (for air at rm temp, 1 atm) a<sub>drag</sub> = 0.5CρAv<sup>2</sup>/m Taking the worse case from the regulations: a<sub>drag,initial</sub> = 0.5(0.5)(1.25 kg/m<sup>3</sup>)(0.00439 m<sup>2</sup>)(27.0 m/s)<sup>2</sup>/(0.142 kg) = 7.04 m/s<sup>2</sup> At 60'6" = 18.4 m, the ball would take a little under a second to reach the plate. Therefore, it can be approximated that the ball will experience only a slight decrease in this deceleration over the flight, so I will estimate that a<sub>drag,final</sub> ~< a<sub>drag,initial</sub>. Rephrasing these estimations: t = 0.70 s, and a<sub>drag</sub> = 6.8 m/s<sup>2</sup>. This gives v<sub>final</sub> = 22 m/s. That is an 20% (substantial) decrease in velocity. In summary: I stand corrected.
