Question

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

• 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 the maximum height attained by the particle be $H$.

Time taken by it to reach maximum height $T=\sqrt{\frac{2H}{g_{eff}}}$

During upward motion :

The particle experiences acceleration due to gravity $10m/s^2$ in downward direction and also, resistive retardation $2m/s^2$ in downward direction.

So, effective acceleration $g_{eff}=10+2=12m/s^2$

Thus, time of ascent $t_a=\sqrt{\frac{2H}{12}}$

During downward motion :

The particle experiences acceleration due to gravity $10m/s^2$ in downward direction and also, resistive retardation $2m/s^2$ in upward direction.

So, effective acceleration $g_{eff}=10-2=8m/s^2$

Thus, time of descent $t_d=\sqrt{\frac{2H}{8}}$

Thus $\frac{t_a}{t_d}=\sqrt{\frac{2}{3}}$

Correct answer is option B.