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. $\frac{2}{3}m$
• C. $\frac{2}{9}m$
• D. $18m$

Answer 1

The correct option is D

$18m$

Step 1: Given data and assumptions

Let the acceleration due to gravity at planet A is $g_A$ and acceleration due to gravity at planet B is $g_B$.

It is given that,

Planet A's gravitational acceleration is 9 times that of planet B's gravitational acceleration. So,

$g_A=9g_B----\left(1\right)$

Step 2: Formula Used:

The third equation of motion is $v^2=u^2+2gh$

Where$V$ is the final velocity,

$u$ is the initial velocity,

$h$ is the height.

Step 3: Determine the height of the jump

We know that the third equation of motion

Initial velocity is zero, so,

$v^2=2gh$.

Let us suppose the height of the jump on planet A is $h_A$

The height of the jump on planet B is $h_B$.

For the planet $A,h_A=\frac{v^2}{2g_A}----\left(2\right)$

For the planet $B,h_B=\frac{v^2}{2g_B}----\left(3\right)$

Divide equation $\left(2\right)$ by equation $\left(3\right)$,

$$\begin{gathered} \frac{h_A}{h_B}=\frac{2g_B}{2g_A} \\ \frac{h_A}{h_B}=\frac{g_B}{9g_B} \\ {h}_B=9h_A \\ {h}_B=9\times 2 \\ {h}_B=18m \end{gathered}$$(From equation 1)

Therefore, the height of the jump on the planet $B$ will be $18m$.

Hence, option D is the correct answer.