A Baseball Player

Classical Mechanics Level 3

A pitcher throws a baseball towards home plate, a distance of $18 \text{ m}$ away, at $v = 40 \text{ m}/\text{s}$ . Suppose the batter takes $.2 \text {s}$ to react before swinging. In swinging, the batter accelerates the end of the bat from rest through $2 \text{ m}$ at some constant acceleration $a$ . Assuming that the end of the bat hits the ball if it crosses the plate within $. 05 \text{ s}$ of the ball crossing the plate, what is the minimum required $a$ in $\text{m}/\text{s}^2$ to the nearest tenth for the batter to hit the ball?

The answer is 44.4.

1 solution

Matt DeCross
May 1, 2016

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:

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

where is 0.3 s as a maximum time to swig the bat coming from?

Soloh Sattoh - 3 years, 9 months ago

.5 s is the longest the batter can wait to make contact, but the batter takes .2 s to just react, leaving 0.5-0.2 = 0.3 s to actually swing.

Matt DeCross - 3 years, 8 months ago

0.5 - 0.2 = 0.3

Ani B - 2 years, 5 months ago

Don't you need to clarify that x-component of velocity is also 40 m/s since there is no initial y velocity, your solution only works assuming there was 0 initial y-velocity otherwise we would need the angle.

Alston Lee - 1 year, 9 months ago

A Baseball Player

The answer is 44.4.

1 solution

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