Question

If an astronaut on the moon drops a hammer from a height of $1.4m$. How long will it take the hammer to reach the ground if the gravitational acceleration on the moon is one-sixth of that on Earth?

Answer 1

Hint: The force of gravity of a body depends on its mass and distance from the centre. Since, the force of gravity on the moon is less than that on Earth; the acceleration due to gravity will also have a smaller value. Using the equations of motions and substituting the corresponding values, we can calculate the time taken by the hammer to reach the ground.
Formulas used:
$v^2=u^2+2as$
$v=u+at$

Complete answer:
The gravitational force on the moon is less than that on the Earth, hence the acceleration due to gravity is less than that on Earth. Let the acceleration due to gravity on the earth be $g$ the, the acceleration due to gravity on the moon will be ${\displaystyle \frac{g}{8}}$.
When the hammer is dropped, only the force of gravity acts on it, so we can use the equations of motion to analyse its motion. Therefore, using the following equation of motion,
$v^2=u^2+2as$
Here, $v$ is the final velocity
$u$ is the initial velocity
$a$ is the acceleration of the body
$s$ is the displacement travelled
Given, $a={\displaystyle \frac{g}{8}}$, $u=0$, $s=1.4m$. We substitute given values in the above equation to get,
$\begin{array}{rl}& v^2=0^2+2\times {\displaystyle \frac{g}{8}}\times 1.4\\ & \Rightarrow v^2=2\times {\displaystyle \frac{10}{8}}\times 1.4\\ & \Rightarrow v\approx 1.9m{s}^{-1}\end{array}$
Therefore the final velocity of the hammer is $1.9m{s}^{-1}$. Now using the following equation of motion given by-
$v=u+at$
Here, $t$ is the time taken
We substitute given value n the equation given above to get,
$\begin{array}{rl}& 1.9=0+{\displaystyle \frac{10}{8}}\times t\\ & \Rightarrow t={\displaystyle \frac{1.9\times 8}{10}}\\ & \Rightarrow t=1.52s\end{array}$
The time taken by the hammer is $1.52s$.
Therefore, the time taken by the hammer to reach the ground is $1.52s$.

Note:
The mechanical energy of the hammer is conserved therefore, the highest potential energy is equal to the highest kinetic energy. Just when the hammer touches the ground, its kinetic energy is highest. Gravitational force is a conservative force, i.e. it depends on the initial and final states of the body and not on the path taken.