An outfielder throws a baseball from the outfield to the third baseman. In flight, the ball reaches a height of 25 ft above the ground. The outfielder releases the ball from a height of 6 ft above the ground with a speed of 124 ft/sec. The ball is caught 6 ft above the ground by the third baseman. Determine the horizontal distance between the point where the ball is released and where it is caught
physics - drwls, Monday, November 22, 2010 at 12:51pm
The ball actually rises H = 19 feet, and falls an equal distance. The travel time of the baseball is twice the time it takes to rise or fall 19 feet. That total time is 2*sqrt[H/(2g)]= 1.086 seconds
The vertical velocity component when thrown is (0.543s)*g = 17.5 ft/s
The horizontal velocity component is (and remains) sqrt[124^2 - 17.5^2] = 122.8 ft/s
That should be multiplied by the flight time to get the distance of the throw.
physics - sandhu, Monday, November 22, 2010 at 8:59pm