## In This post you will get answer of Q.

The acceleration due to gravity on the planet $A$ is $9$ times the acceleration due to gravity on planet $B. A$ man jumps to a height of $2, m$ on the surface of $A.$ What is the height of jump by the same person on the planet $B$.

Solution:

## It is given that, acceleration due to gravity on plane A is 9 times the acceleration due to gravity on planet B ie,

$g_A=9g_B$ …….(i)

From third equation of motion

$v^2=2gh$

At planet A, $h_{A}=frac{v^2}{2g_{A}}$ ……(ii)

At planet B, $h_{B}=frac{v^2}{2g_{B}}$ ……(iii)

Dividing Eq. (ii) by Eq. (iii), we have

$frac{h_{A}}{h_{B}}=frac{g_{B}}{g_{A}}$

From Eq. (i), $g_A = 9,g_B$

= $frac{h_{A}}{h_{B}}=frac{g_{B}}{9g_{B}}=frac{1}{9}$

or $h_B=9,h_{A}=9times2,=18,m (=h_{A}=2,m)$