Question

A particle is thrown upwards from ground. It experiences a constant air resistance force which can produce a retardation of $2m/s^2.$ The ratio of time of ascent to the time of descent is $:[g=10m/s^2]$

• A. $1:1$
• B. $\sqrt{\frac{2}{3}}$
• C. $\frac{2}{3}$
• D. $\sqrt{\frac{3}{2}}$

Answer 1

The correct option is B $\sqrt{\frac{2}{3}}$

Let a be the retardation produced by resistive force, $t_{a}and{t}_d$ be the time ascent and descent respectively.
Let the particle rise upto a height $h$ and projection velocity be $u$.

For upward motion,

$0=u-(a+g)t_a..........\left(i\right)$

$0=u^2-2(a+g)h..........\left(ii\right)$

For downward motion,

$v_2=0+(g-a)t_d.........\left(iii\right)$

$v_2^2=0+2(g-a)h.......\left(iv\right)$

From $\left(ii\right)\&\left(iv\right),$

$v_2=u_2\sqrt{\frac{g-a}{g+a}}...........\left(v\right)$

From $\left(i\right)\&\left(iii\right),$

$\frac{t_a}{t_d}=\frac{u(g-a)}{v_2(g+a)}............\left(vi\right)$

From $\left(v\right)\&\left(vi\right),$

$\frac{t_a}{t_d}=\sqrt{\frac{g-a}{g+a}}$

$=\sqrt{\frac{2}{3}}$