A batter hits the ball along the 1st baseline, with an initial velocity of 60.0

Question

A batter hits the ball along the 1st baseline, with an initial velocity of 60.0 mph @ an angle of 73 degrees with respect to the ground. The rightfielder is located 180 ft from homeplate, at an angle of 30 degrees from the 1st baseline. After a delay of 0.312 seconds, the rightfielder starts moving in a straight line towards the ball's estimated landing location. He can accelerate from rest to a top speed of 8.5 m/s in 6.90 seconds. (a)Where does the ball land? ____m at an angle of _____ degrees off the first base line. (b) Does he catch the ball? Yes No Not Enough Information ***Assume that: 1) The batter hits the ball just above the ground level 2) YOu can ignore the physical dimensions of the right fielder 3) AND, IF the right fielder catches it, he catches it just before he lands.

Explanation / Answer

Vertical velocity (uv) = 60 * sin(73) = 57.378 mph horizontal velocity (uh)= 17.54 mph total time for which ball is in air = 2*uv/g = 2*57.378*1.6*1000/(9.8*3600) = 5.204 sec i.e horizontal distance by ball in this duration = 17.54*1.6*1000*5.204/3600 = 40.568 m i.e ball lands at an angle of 73 degree with ground (as parabolic trajectory) at a distance of 40.568 m from the place of batter. Catcher needs to cover 180 *0.3048 - 40 = 14.864m catcher has acceleration as 8.5/6.9 = 1.232 so he has total time = 5.204 - 0.312= 4.892s i.e. maximum distance possible in this time = 0.5* 1.232 * (4.892^2) = 14.74m Hence, catcher just misses this catch!!

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