Tuesday, May 20, 2014

Tracker: rachelong student video tennis ball drop in water motion

Tracker: rachelong student video tennis ball drop in water motion

physics of tennis ball drop in water
Kinematics Model
y = if ( t<0.29,(-2.137E0)*t^2-1.976E-1*t+2.310E-1,if(t<0.735,9.542E-1*(t)^2-9.368E-1*(t)+0.17,if(t<0.902,-9.661E-1*t^2+1.592E0*t-6.479E-1,if(t<1.169,6.172E-1*t^2-1.284E0*t+6.580E-1,0))))
author of video: rachelong
author of model: lookang

first part of the kinematics model y = (-2.137E0)*t^2-1.976E-1*t+2.310E-1
in physics, it means it is free falling under gravity with fy = 2.137*2 = 4.274 m/s^2 assuming m = 1kg, no noticeable air resistance.

second part of the kinematics model y = 9.542E-1*(t)^2-9.368E-1*(t)+0.17
the physics here is fy = mg - upthrust = m*ay, in simple words, it is the upthrust that causes the eventual motion in the water as long as the ball is completely immersed. the data of the video analysis suggests the resultant ay = 0.9542*2 = 1.9084 m/s^2 due to upthrust ( $\row g y $) minus the weight (mg)

third part of the kinematics model y = -9.661E-1*t^2+1.592E0*t-6.479E-1
the physics mostly fy = mg neglecting water droplets and bouncing if any from surfacing out the water surface.
the ay = -0.9661*2 = - 1.9322 m/s^2

four part of the kinematics model y = 6.172E-1*t^2-1.284E0*t+6.580E-1
the tennis ball is back under water, assuming the upthrust and weight now acts on the ball.
though it is not determine though the video, what is the upthrust and weight values, we can determine the resultant ay as a result of the 2 forces. ay = 0.6172*2 =1.2344 m/s^2.


  1. why is the value of ay determined in part one or first part not equal to -9.81 m/s^2? is the frame-rate changed accidentally during compression process in iMovie software?
  2. can you explain the motion using the following kinematics model y = if ( t<0.29,(-2.137E0)*t^2-1.976E-1*t+2.310E-1,if(t<0.735,9.542E-1*(t)^2-9.368E-1*(t)+0.17,if(t<0.902,-9.661E-1*t^2+1.592E0*t-6.479E-1,if(t<1.169,6.172E-1*t^2-1.284E0*t+6.580E-1,0)))) ?