A baseball being hit is in contact with the bat for 1.30 ms. The ball is moving toward the bat at speed 38.0 m/s just before it is hit, and is moving at 67.5 m/s away from the bat just after it is hit. Assume the ball has a constant acceleration while it is in contact with the bat, and that the ball’s motion is always along a straight line.
(a)Calculate the acceleration of the ball while it is being hit.
(b)What is the ball’s net displacement while it is in contact with the bat?
a)Constant acceleration a reduces initial velocity and increases inverse velocity for t = 1.3*10¯³ s. General change of velocity is Δν=(38+67.5) m/s, hence: Δν=at a=Δν/t=((38+67.5)m/s)/1.3*10¯³s≈81.154*10³ m/s² b)Ball’s net displacement is: Δx=(νο*t)+(at²)/2=38m/s*1.3*10¯³s+(81.154*10³ m/s²*(1.3*10¯³s)²/2)≈0.118m=11.8cm