Question

A stone thrown vertically upward, takes $3s$ to attain the maximum height. Calculate the maximum height attained by the stone.

[Take $g=9.8m{s}^{-2}$]

Answer 1

Step 1: Given data

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

Time taken, $t=3s$

Acceleration due to gravity, $g=9.8m{s}^{-2}$

Final velocity of the stone, $v=0$

Height attained by the stone, $h=?$

Step 2: Calculating the initial velocity of the stone

Applying equation of motion for the freely falling object,

$$\begin{gathered} v=u+gt \\ \Rightarrow 0=u+(-9.8m{s}^{-2})\times 3s \\ \Rightarrow u=29.4m{s}^{-1} \end{gathered}$$ [ '$g$' is negative when a body is projected upwards]

Step 3: Calculating the maximum height attained by the stone

Applying equation of motion for the freely falling object,

$$\begin{gathered} v^2=u^2+2gh \\ \Rightarrow 0={(29.4m{s}^{-1})}^2+2\times (-9.8m{s}^{-2})\times h \\ \Rightarrow h=44.1m \end{gathered}$$ [ '$g$' is negative when a body is projected upwards]

Hence, the maximum height attained by the stone is $44.1m$