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Question

Wood bats are required in professional baseball but metal bats are sometimes allowed in youth and college baseball. One result is that the exit speed vv of the baseball off a metal bat can be greater. In one set of measurements under the same circumstances, v=50.98 m/sv=50.98 \mathrm{~m} / \mathrm{s} off a wood bat and v=61.50 m/sv=61.50 \mathrm{~m} / \mathrm{s} off a metal bat. Consider a ball hit directly toward the pitcher. The regulation distance between pitcher and batter is Δx=60ft6in\Delta x=60 \mathrm{ft} 6 \mathrm{in}. For those measured speeds, how much time Δt\Delta t does the ball take to reach the pitcher for (a)(a) the wood bat and (b)(b) the metal bat? (c)(c) By what percentage would Δt\Delta t be reduced if professional baseball switched to metal bats? Because pitchers do not wear any protective equipment on face or body, the situation is already dangerous and the switch would add to that danger.

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Givens

  • vw=50.98 msv_w=50.98\ \frac{\text{m}}{\text{s}} - the exit speed off a wood bat
  • vm=61.5 msv_m=61.5\ \frac{\text{m}}{\text{s}} - the exit speed off a metal bat
  • l=60 ft 6 in=18.4 ml=60\ \text{ft}\ 6\ \text{in}=18.4\ \text{m} - the distance between pitcher and batter

Required

  • a) Δtw\Delta t_w - the time it takes for the wood bat
  • b) Δtm\Delta t_m - the time it takes for the metal bat
  • c) kk - the reduction in the time it takes

How can we use the known expression for the distance travelled by the constant speed ball in order to solve this problem?

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