Question

14) A particle, falling under gravity, falls 20 metres in a certain second. The time required to cover next 20 metres will be

Answer 1

Let $\sf{g=10\ m\,s^{-2}.}$

Since the particle covers a distance of 20 metres in a certain second (t = 1 s), by second equation of motion,

$\displaystyle\longrightarrow\sf{s=ut+\dfrac{1}{2}\,at^2}$

$\displaystyle\longrightarrow\sf{20=u\times1+\dfrac{1}{2}\times10\times1^2}$

$\displaystyle\longrightarrow\sf{20=u+5}$

$\displaystyle\longrightarrow\sf{u=15\ m\,s^{-1}}$

After travelling this 20 metres, the velocity of the particle will be, by first equation of motion,

$\displaystyle\longrightarrow\sf{v=u+at}$

$\displaystyle\longrightarrow\sf{v=15+10\times1}$

$\displaystyle\longrightarrow\sf{v=15+10}$

$\displaystyle\longrightarrow\sf{v=25\ m\,s^{-1}}$

Hence the time required to cover the next 20 metres is given by the second equation of motion as $\sf{(u=25\ m\,s^{-1}),}$

$\displaystyle\longrightarrow\sf{s=ut+\dfrac{1}{2}\,at^2}$

$\displaystyle\longrightarrow\sf{20=25t+\dfrac{1}{2}\times10t^2}$

$\displaystyle\longrightarrow\sf{20=25t+5t^2}$

$\displaystyle\longrightarrow\sf{5t^2+25t-20=0}$

$\displaystyle\longrightarrow\sf{t^2+5t-4=0}$

$\displaystyle\longrightarrow\sf{t=\dfrac{-5+\sqrt{5^2-4\times1\times-4}}{2\times1}\quad\quad[t\geq0]}$

$\displaystyle\longrightarrow\sf{t=\dfrac{\sqrt{41}-5}{2}\ s}$

$\displaystyle\longrightarrow\sf{\underline{\underline{t=0.702\ s}}}$