Question

A stone thrown vertically upward rises to a height of $29.34m$ on the surface of the Moon. If the acceleration due to gravity on the surface of the Moon is $1.63m{s}^{-2}$. Calculate the time taken to gain the maximum height.

Answer 1

Step 1: Given data

Initial velocity, $u=?$ (To be calculated)

Final velocity of the stone, $v=0$

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

Height attained by the stone, $h=29.34m$

Step 2: Calculating the initial velocity of the stone

Applying the equation of motion for the freely falling object,

$$\begin{gathered} v^2=u^2+2gh \\ \Rightarrow 0=u^2+2\times (-1.63m{s}^{-2})\times 29.34m \\ \Rightarrow u=9.78m{s}^{-1} \end{gathered}$$ [ '$g$' is negative when a body is projected upwards]

Step 3: Calculating the time taken to gain the maximum height

$$\begin{gathered} v=u+gt \\ \Rightarrow 0=9.78m{s}^{-1}+(-1.63m{s}^{-2})\times t \\ \Rightarrow t=\frac{9.78m{s}^{-1}}{1.63m{s}^{-2}} \\ \Rightarrow t=6s \end{gathered}$$ [ '$g$' is negative when a body is projected upwards]

Hence, the time is taken to gain the maximum height is $6s$.