1.

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»

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


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