Question

Two bodies are held separated by 9.8m vertically one above the other . They are released simultaneously to fall freely under gravity . After 2s the relative distance between them is?

Answer 1

Given:

Two bodies are held separated by 9.8m vertically one above the other . They are released simultaneously to fall freely under gravity.

To find:

Relative distance after 2 seconds ?

Calculation:

Displacement of lower body after 2 seconds:

$d1 = 9.8 + (ut + \dfrac{1}{2} g {t}^{2} )$

$\implies d1 = 9.8 + (0 + \dfrac{1}{2} g {t}^{2} )$

$\implies d1 = 9.8 + \dfrac{1}{2} g {(2)}^{2}$

Now, displacement of upper body after 2 seconds:

$d2= ut + \dfrac{1}{2} g {t}^{2}$

$\implies d2 = 0 + \dfrac{1}{2} g {t}^{2}$

$\implies d2 = \dfrac{1}{2} g {(2)}^{2}$

Now, relative displacement between them:

$d = d1 - d2$

$\implies \: d = \{ 9.8 + \dfrac{1}{2} g {(2)}^{2} \} - \dfrac{1}{2} g {(2)}^{2}$

$\implies \: d = 9.8 \: m$

So, relative displacement after 2 sec is 9.8 metres.