Question

The value of g on the moon in 1/6 th of the value of g on earth. A man can jump 1.5 m high on the earth. He can jump on the moon up to a height of:
A.) 9 m
B.) 7.5 m
C.) 6 m
D.) 4.5 m

Answer 1

Hint: Since the gravity in the moon is low compared to the earth, the slight movement can make a large impact. You can use laws of motions to find the maximum height obtained on the moon.

Complete step by step answer:
According to the law of motions,
$$v^2=u^2+2as$$, where v is the final velocity, u is the initial velocity, a is the acceleration and s is the displacement.
In this case the final velocity will be zero, displacement will be the maximum height and acceleration will be negative since we are acting against the gravity.
$$0=u^2-2g{h}_{max}$$
$$u^2=2g{h}_{max}$$
Hence the maximum height, $$h_{max}={\displaystyle \frac{u^2}{2g}}$$
From this equation we can see that the maximum height is inversely proportional to the acceleration due to gravity.
Let the acceleration due to gravity on earth is $$g_e$$ and the acceleration due to gravity on the moon is $$g_m$$. Maximum height achieved in the earth is $$h_e$$ and maximum height can be achieved in the moon is $$h_m$$.
Since $$h_{max}\propto {\displaystyle \frac{1}{g}}$$, we can compare the maximum height that can be achieved on the moon and earth.
So the ratio will be, $${\displaystyle \frac{h_{\mathrm{m}}}{h_e}}\propto {\displaystyle \frac{g_e}{g_m}}$$
We can assign given values into the equation.
$${\displaystyle \frac{h_{\mathrm{m}}}{1.5}}\propto {\displaystyle \frac{1}{{}^1/{}_6}}$$
Equate the values to get maximum height on the moon.
$$h_m=6\times 1.5$$
$$h_m=9m$$
Hence the option (A) is correct.

Note: Candidates are advised to put the gravity values correctly. If you put wrongly, the ratio will provide a wrong answer to the question. Generally, you can make an assumption that the jump in the moon will be larger than the jump generated in the earth. It is better to remember the gravity relation of the moon compared to the earth.