Posted by sandhu on Monday, November 22, 2010 at 8:44am.
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
Thanks