Question

A particle falling under gravity falls 20 m in a certain second find the time required to cover that next 20 metres

Answer 1

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

The particle covers a distance of 20 metres in a certain second (t = 1 s). So 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 for 1 second, the velocity of this particle will be, by first equation of motion,

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

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

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

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

This is the initial velocity when the particle covers the next 20 metres.

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

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

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

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

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

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

Since $\sf{t\geq 0,}$

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

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

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