Question

A particle is projected vertically up from ground with speed u. Let air applies a resistive force equal to half of particle's weight in opposite of particle's instantaneous velocity. What is the ratio of time of ascent to time of descent?

Answer 1

Retardation is always opposite to velocity.

For upward motion the effective retardation g' is (g+2)=12 m/s^2.

Therefore , t( up) = u/g'=u/12. Here, u is initial velocity……………(1)

The height that will be reached =h= u^2/(2 g')=u^2/24………..(2)

The particle starts moving downward from height h with acceleration g"=10–2=8m/s^2.

Therefore, time of descend t( down)= (2h/g")^1/2= [ 2(u^2/24)/8]^1/2 =u/[4xsqrt 6]…….(3).

From (1) and (2), t(up)/t(down)=(u/12)/u/[4x sqrt 6]=(2/3)^1/2.