Question

A stone dropped from a cliff on the surface of the Moon, reaches the surface of the moon in $30s$. If the acceleration due to gravity of the Moon is $1.63m{s}^{-2}$, calculate the height of the cliff?

Answer 1

Step 1: Given data

Initial velocity of stone, $u=0$

Time taken, $t=30s$

Acceleration due to gravity, $g=1.63m{s}^{-2}$

Height, $h=?$ (To be calculated)

Step 2: Formula used

$h=ut+\frac{1}{2}g{t}^2$ where h is the height, u is the initial velocity, g is the acceleration due to gravity and t is the time,

Step 3: Calculating the height of the cliff

Applying equation of motion for the freely falling object,

$$\begin{gathered} h=ut+\frac{1}{2}g{t}^2 \\ \Rightarrow h=0\times 30s+\frac{1}{2}\times 1.63m{s}^{-2}\times {(30s)}^2 \\ \Rightarrow h=733.50m \end{gathered}$$ [ '$g$' is positive when a body falls downwards]

Hence, the height of the cliff is $733.50m$