On a frozen rope from third to first base is 127 feet 3 and 3/8 inches.
The throw from first to third is 127' 3-3/8 inches.
That would depends on where they field the ball. Howeverthe distance from the 3rd base bag and the 1st base bag is just under 85 feet.
To calculate the distance the outfielder would have to throw the ball to first base, we can use the Pythagorean theorem. The distance from the outfielder to third base is 40 feet, and the distance from third base to first base is 90 feet. So, the total distance the outfielder would have to throw the ball can be calculated as the square root of (40^2 + 90^2), which equals approximately 97.9 feet. Therefore, the outfielder would have to throw the ball approximately 97.9 feet to reach first base.
The distance between first base and second base is 90 feet.
127 feet, 3 3/8 inches
When hit by right hand-ers (the majority of batters) it means the ball was pulled that direction and is going much faster. So the 3rd baseman must have fast reflexes. Also the 3rd baseman then has to make the throw to 1st base. This is a far throw and the 3rd baseman must have a strong accurate arm.
127 feet
120ft. and 6in.
it's 63.72 ft
It's 127 feet, 3 3/8 inches from home plate to second base (the same distance from first base to third base).
That depends on where the 1st baseman fields the ball. If the 1st baseman can field the ball, throw to second, and have enough time to get safely back to the base to catch the return throw, then (s)he will cover. If not, then it is the responsibility of the pitcher.
The distance from home plate to first base is 90 feet and the distance from first base to second base is also 90 feet making a right angle; you can calculate how far the catcher needs to throw to 2nd base from home by Pythagorean theorem. Answer is 127.3 feet