The answer is 44.4.

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The ball takes

$t = \frac{d}{v} = \frac{18 \text{ m}}{40 \text{ m}/\text{s}} = .45 \text{ s}$

to reach the plate, but the batter has until $.45 \text{ s} + .05 \text{ s} = .5 \text{ s}$ to make contact with it due to the buffer.

Since the batter takes $\SI{0.2}{\second}$ to react, we have a maximum time of $t_* = \SI{0.3}{\second}$ to swing the bat.

The distance the end of the bat has traveled as a function of its acceleration and time is $d = \frac12 at^2$ , so to reach home plate in $t_*$ requires an acceleration of:

$a = \frac{2d}{t_*^2} = \frac{4 \text{ m}}{(.3 \text{ s})^2} = 44.\bar{4} \text{ m}/\text{s}^2.$