Question

A boy jumps a distance of 2m on the surface of the earth. What distance will he jump on the surface of the moon where g is 1/6th of the value on the surface of the earth?

Answer 1

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

Answer 2

Answer:

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

g

m

=

6

1

⇒ g_e= 6g_mg

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}

2g

e

u

2

⇒ 2 = \frac{ u^{2} }{2g_e}

2g

e

u

2

putting g_e= 6g_mg

e

=6g

m

⇒ 2 = \frac{ u^{2} }{12g_m}

12g

m

u

2

⇒12 = \frac{ u^{2} }{2g_m}

2g

m

u

2

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