Question

At some planet \['g '\] is \[1.96\,m/{s^2}\]. If it is safe to jump from a height of 2 m on earth, then what should be the corresponding safe height for jumping on that planet?
A. 5 m
B. 2 m
C. 10 m
D. 20 m

Answer 1

Hint: Use the formula for gravitational potential energy. The gravitational potential energy on that planet will be equal to the gravitational potential energy on the earth.

Formula used:

The gravitational potential energy,
\[U = mgh\]

Where, m is the mass, g is the acceleration due to gravity, and h is the height.

Complete step by step answer:
The quantity that relates both acceleration due to gravity and height above the ground is gravitational potential energy. Since it is safe to jump from respective heights on both earth and that planet, the gravitational potential energy must be the same.

Therefore, we can write,
\[m{g_E}{h_E} = m{g_P}{h_P}\]

Here, m is the mass of the person, \[{g_E}\] is the acceleration due to gravity of earth, \[{h_E}\] is the safe height on earth, \[{g_P}\] is the acceleration due to gravity on that planet and \[{h_P}\] is the safe height on that planet.

Rearrange the above equation for \[{h_P}\] as follows,
\[{h_P} = \dfrac{{{g_E}{h_E}}}{{{g_P}}}\]

Substitute \[9.8\,m/{s^2}\] for \[{g_E}\], 2 m for \[{h_E}\] and \[1.96\,m/{s^2}\] for \[{g_P}\] in the above equation.
\[{h_P} = \dfrac{{\left( {9.8\,m/{s^2}} \right)\left( {2\,m} \right)}}{{1.96\,m/{s^2}}}\]
\[ \Rightarrow {h_P} = \dfrac{{19.6\,{m^2}/{s^2}}}{{1.96\,m/{s^2}}}\]
\[ \Rightarrow {h_P} = 10\,m\]

Therefore, the safe height for jumping on that planet will be 10 m as the acceleration gravity is very less as compared to acceleration due to gravity on the surface of earth.

So, the correct answer is option (C).

Note: We can also answer this question by taking the ratio of acceleration due to gravity on that planet to the acceleration due to gravity on the earth. The acceleration due to gravity on that planet will be \[{g_P} = \dfrac{{{g_E}}}{5}\]. Therefore, the safe height for jumping on that planet would be 5 times the safe height on the earth.