Question

The acceleration due to gravity on planet A is 9 times the acceleration due to gravity on planet B. A man jumps to a height of 2m on the surface of A. What is the height of the jump by the same person on planet B?
(A) $6m$
(B) $\dfrac{2}{3}m$
(C) $\dfrac{2}{9}m$
(D) $18m$

Answer 1

Hint: In order to solve this question, we should know the third equation of motion. According to the third equation of motion, we have ${v^2} - {u^2} = 2as$. Here, $u$is the initial speed of the body. v is the final speed of the body. $a$is the acceleration of the body and $s$is the distance covered by the body.

Complete answer:
Given that the acceleration due to gravity on planet $A$is $9$times the acceleration due to gravity on planet$B$. It can be represented in the following way:
${g_A} = 9{g_B}{\text{ }}...{\text{(1)}}$
We have been asked for the height of the jump by the same person on planet B.

From the third equation of motion, we have the following equation:
${v^2} = 2gh{\text{ }}...{\text{(2)}}$
At planet$A$, we have ${h_A} = \dfrac{{{v^2}{\text{ }}}}{{2{g_A}}}{\text{ }}...{\text{(3)}}$
At planet$B$, we have ${h_B} = \dfrac{{{v^2}{\text{ }}}}{{2{g_B}}}{\text{ }}...{\text{(4)}}$

Now, divide equation $(3)$ by equation$(4)$.
$ \Rightarrow \dfrac{{{h_A}}}{{{h_B}}} = \dfrac{{{g_B}{\text{ }}}}{{{g_A}}}$
$ \Rightarrow \dfrac{{{h_A}}}{{{h_B}}} = \dfrac{{{g_B}{\text{ }}}}{{9{g_B}}}$
$ \Rightarrow \dfrac{{{h_A}}}{{{h_B}}} = \dfrac{1}{9}$
$ \Rightarrow {h_B} = 9{h_A}$
$ \Rightarrow {h_B} = 9 \times 2$
$ \Rightarrow {h_B} = 18~m$
So, the height of the jump by the same person on planet B is 18m.

Therefore, the correct option is (D) 18 m.

Note: According to Newton’s third law of motion, if an object exerts a force on another object, then that object must exert a force of equal magnitude and opposite direction back on the first object. This law of motion signifies a specific symmetry in nature that forces always occur in pairs and one body cannot exert a force on the other body without experiencing a force itself.