Answer:
18 m
Explanation:
G = Gravitational constant
m = Mass of planet = [tex]\rho V[/tex]
[tex]\rho[/tex] = Density of planet
V = Volume of planet assuming it is a sphere = [tex]\dfrac{4}{3}\pi r^3[/tex]
r = Radius of planet
Acceleration due to gravity on a planet is given by
[tex]g=\dfrac{Gm}{r^2}\\\Rightarrow g=\dfrac{G\rho V}{r^2}\\\Rightarrow g=\dfrac{G\rho \dfrac{4}{3}\pi r^3}{r^2}\\\Rightarrow g=\dfrac{4G\rho\pi r}{3}[/tex]
So,
[tex]g\propto \rho r[/tex]
Density of other planet = [tex]\rho_p=\dfrac{1}{4}\rho_e[/tex]
Radius of other planet = [tex]r_p=\dfrac{1}{3}r_e[/tex]
[tex]\dfrac{g_e}{g_p}=\dfrac{\rho_e r_e}{\rho_p r_p}\\\Rightarrow \dfrac{g_e}{g_p}=\dfrac{\rho_e r_e}{\dfrac{1}{4}\rho_e\times \dfrac{1}{3}r_e}\\\Rightarrow \dfrac{g_e}{g_p}=12\\\Rightarrow g_p=\dfrac{g_e}{12}\\\Rightarrow g_p=\dfrac{9.8}{12}[/tex]
Since the person is jumping up the acceleration due to gravity will be negative.
From kinematic equations we have
[tex]v^2-u^2=2g_es\\\Rightarrow u^2=v^2-2g_es\\\Rightarrow u^2=0-2\times -9.8\times 1.5\\\Rightarrow u^2=2\times 9.8\times 1.5[/tex]
On the other planet
[tex]v^2-u^2=2g_ps\\\Rightarrow s=\dfrac{v^2-u^2}{2g_p}\\\Rightarrow s=\dfrac{0-(2\times 9.8\times 1.5)}{2\times -\dfrac{9.8}{12}}\\\Rightarrow s=18\ \text{m}[/tex]
The man can jump a height of 18 m on the other planet.
