# A red and a green stone falls freely next to each other along vertical paths at a place of the universe where the acceleration due to gravity is exactly 10 m/s^2 and where nothing prevents free fall. The red stone just starts falling when the ......?

## A red and a green stone falls freely next to each other along vertical paths at a place of the universe where the acceleration due to gravity is exactly $10 \frac{m}{s} ^ 2$ and where nothing prevents free fall. The red stone just starts falling when the green stone reaches it. The green stone also started from rest but from a higher position. After a while the distance between the two stones is $7 m$, and then after two more seconds elapses the distance between them will increase to $27 m$ . How much higher did the green stone start in $m$?

Mar 19, 2017

$25 m$

#### Explanation:

$s = u t + g {t}^{2}$ .....(1)
$v = u + g t$ ......(2)

1. Let green stone start from rest from a height ${h}_{g}$ above red stone. Time taken by green stone to cross red stone $= {t}_{1 g}$ We get
${h}_{g} = 0 \times {t}_{1 g} + \frac{1}{2} \times 10 \times {t}_{1 g}^{2}$
$\implies {t}_{1 g} = \sqrt{{h}_{g} / 5}$ .....(3)
2. From (2), Velocity of green stone as it crosses red stone
${v}_{1 g} = 0 + 10 \times {t}_{1 g}$
$\implies {v}_{1 g} = 10 {t}_{1 g}$
3. It is clear that green stone will always be in the lead.
Let after $t s$ of green stone crossing the red stone, distance between the two is $7 m$
${s}_{g} - {s}_{r} \left(\overline{t}\right) = \left(10 {t}_{1 g} t + 10 {t}^{2}\right) - \left(0 \times t + 10 {t}^{2}\right) = 7$
$\implies 10 {t}_{1 g} t = 7$

Using (3), we get
$10 \sqrt{{h}_{g} / 5} t = 7$
$\implies t = \frac{7}{10} \sqrt{\frac{5}{h} _ g}$ ......(4)
4. It is given that after $\left(t + 2\right) s$ distance between the stones becomes $27 m$
Again using (1) we get
${s}_{g} - {s}_{r} \left(\overline{t + 2}\right) = \left(10 {t}_{1 g} \left(t + 2\right) + 10 {\left(t + 2\right)}^{2}\right) - \left(0 \times \left(t + 2\right) + 10 {\left(t + 2\right)}^{2}\right) = 27$
$\implies 10 {t}_{1 g} \left(t + 2\right) = 27$

Using (3) and (4) we get
$10 \sqrt{{h}_{g} / 5} \left(\frac{7}{10} \sqrt{\frac{5}{h} _ g} + 2\right) = 27$
$\implies \left(7 + 2 \times 10 \sqrt{{h}_{g} / 5}\right) = 27$
$\implies 20 \sqrt{{h}_{g} / 5} = 20$
$\implies {h}_{g} = 25 m$