Question
The acceleration due to gravity on the planet $$A$$ is 9 times the acceleration due to gravity on the 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 $$B$$ ?
A.
$$6\,m$$
B.
$$\frac{2}{3}\,m$$
C.
$$\frac{2}{9}\,m$$
D.
$$18\,m$$
Answer :
$$18\,m$$
Solution :
It is given that, acceleration due to gravity on planet $$A$$ is 9 times the acceleration due to gravity on planet $$B$$ i.e.
$${g_A} = 9{g_B}\,......\left( {\text{i}} \right)$$
From third equation of motion, $${v^2} = 2gh$$
At planet $$A,$$ $${h_A} = \frac{{{v^2}}}{{2{g_A}}}\,......\left( {{\text{ii}}} \right)$$
At planet $$B,$$ $${h_B} = \frac{{{v^2}}}{{2{g_B}}}\,......\left( {{\text{iii}}} \right)$$
Dividing Eq. (ii) by Eq. (i), we have
$$\frac{{{h_A}}}{{{h_B}}} = \frac{{{g_B}}}{{{g_A}}}$$
From Eq. (i), $${g_A} = 9{g_B}$$
$$\eqalign{
& \therefore \frac{{{h_A}}}{{{h_B}}} = \frac{{{g_B}}}{{9{g_B}}} = \frac{1}{9} \cr
& {\text{or}}\,\,{h_B} = 9{h_A} = 9 \times 2 = 18\;m\,\,\left( {\because {h_A} = 2\;m} \right) \cr} $$