A batter hits the ball along the 1st baseline, with an initial velocity of 49.0

Question

A batter hits the ball along the 1st baseline, with an initial velocity of 49.0 mph @ an angle of 60 degrees with respect to the ground. The rightfielder is located 106 ft from homeplate, at an angle of 19 degrees from the 1st baseline. After a delay of 0.300 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 7.5 m/s in 6.50 seconds. (a)Where does the ball land? ? m, at an angle of ? degrees off the 1st baseline. <---(how you'll answer) (b)Does he catch the ball? Assumptions:

Explanation / Answer

a) u = 49 mph = 21.91 m/s

uy = 21.91sin60 = 18.97 m/s

in vertical : -uy = uy - gt

t = 2 x 18.97/9.81 = 3.88 sec

horizontally : d = 21.91cos60 x 3.88 =42.5 m at 0 degrees from 1st baseline......Ans

b) assume baseline as x-axis

106ft = 32.31 m/s

distance thats cathcer to have traveled : (42.5 - 32.31cos15 )i + (0 - 32.31sin15)

d = 11.30 i - 8.36 j m

|d| = sqrt(11.30^2 + 8.36^2) = 14.10 m

a = 7.5/6.5 = 1.15 m/s2

time he have = 3.88 - 0.30 = 3.58 s

d = ut + at^2 /2

14.10 = 0 x t + 1.15t^2 /2

t = 4.95 sec

he needs 4.95 sec to cover this distance but he have only 3.58 sec .

so won't be able to catch this ball.

hope this help

thanks

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