Question

The weight of a person on a planet A is about half that on the earth. He can jump up to a height of $0.4m$ on the surface of the earth. How high can he jump on planet A?

Answer 1

Step 1: Given that

The weight of the person on planet A is half that on earth

$W_A=\frac{W_{earth}}{2}$

$W$ is the weight on the planet.

The height the person can jump is $0.4m$

$H_{earth}=0.4m$

$H$ is the height of the jump.

Step 2: Formula used

$$\begin{gathered} W=mg \\ \mathrm{K}.\mathrm{E}=\frac{m{v}^2}{2} \\ \mathrm{P}.\mathrm{E}=mgH \end{gathered}$$

Here,

$m$ is the mass of the object.

$g$ is the acceleration due to gravity on the planet.

$v$ the initial velocity of the pump.

$K.E$ is the kinetic energy.

$P.E$ is the potential energy.

Step 3: Calculating the maximum height of the jump on a planet

Initial kinetic energy =$\frac{m{v}^2}{2}$

Final kinetic energy = $0$ as $v=0$

Initial potential energy = $0$ as $H=0$

Final potential energy = $m{g}_{planet}{H}_{planet}$

Using the law of conservation of energy -

$0+\frac{m{v}^2}{2}=0+m{g}_{planet}{H}_{planet}$

$H_{planet}=\frac{v^2}{2g_{planet}}$

Step 4: Finding the relation between g

$$\begin{gathered} W_A=\frac{W_{earth}}{2} \\ m{g}_A=\frac{m{g}_{earth}}{2} \\ {g}_A=\frac{g_{earth}}{2} \end{gathered}$$

Step 5: Calculating the height of the jump

$$\begin{gathered} h_{earth}=0.4m \\ \frac{v^2}{2g_{earth}}=0.4m \\ \frac{v^2}{4g_A}=0.4m \\ \frac{v^2}{2g_A}=0.8m \\ {h}_A=0.8m \end{gathered}$$

Hence, the person can jump up to 0.8 m on planet A.