First slide
Kinetic friction
Question

Two bodies of different masses are dropped simultaneously from same height. If air friction acting on them is directly proportional to the square of their mass, then ,

Moderate
Solution

f = k.m2 where k = constant.

.: a = mg-f/m = (g-km)

Again , t = √(2h/a)

Therefore , If 'm' increases , 'a' decreases and 't' increases.

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