Answers

2015-02-10T15:30:51+05:30
So we see at the top most point the velocity the boy is zero and the earths pull is acting downwards

Given      \frac{g_m}{g_e}= \frac{1}{6}

⇒ g_e= 6g_m


so using

v² = u² - 2gs            where g = acceleration due to gravity
                                       s = total height

so 

0 = u² - 2gs  

⇒ s =  \frac{ u^{2} }{2g_e}

⇒ 2 =  \frac{ u^{2} }{2g_e}

putting    g_e= 6g_m

⇒  2 =  \frac{ u^{2} }{12g_m}

⇒12 =  \frac{ u^{2} }{2g_m}

so the boy can jump a height of 12 m on moon's surface

2 4 2