A small piece of styrofoam packing material is dropped from a height of 2m above the ground. Until it reached terminal speed, the magnitude of its accelleration is given by a=g-bv. After it fall 0.500m, the foam effectively reaches terminal speed and then takes 5.00 s more to reach the ground.
a) what is the value of the contant b?
b) what s the accelleration when the speed is 0.150m/s?
so far i know the accelleration is 9.8m/s^2 when t=0 .... i simply cannot find the constant ...my dad said its 32.75..but he refuse to tell how he came by this answer...
Thanks in advance
2007-08-26 01:00:59 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Physics
a) At 1.5 m height v = terminal speed, which is 1.5/5 m/s.
At that speed a=0 (no further speed change), so 9.8 = bv, b = 9.8/v. Solve that.
b) Now that you know b, you can solve this part. a = 9.8 - 0.15b.
2007-08-27 02:48:08 · answer #1 · answered by kirchwey 7 · 0⤊ 0⤋