Question

A stone is thrown vertically upward with an initial velocity of $40m{s}^{-1}$. Taking $g=10m{s}^{-2}$, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

Answer 1

Step 1: Given

Initial velocity (u) = $40m{s}^{-1}$ (in upward direction)

Final velocity (v) = $0m{s}^{-1}$

Acceleration due to gravity, $g=10m{s}^{-2}$

Step 2: Formula used and calculation

$$\begin{gathered} v^2=u^2+2as \\ 0={40}^2+2\times (-10)\times s \end{gathered}$$

(-ve sign is used because the stone is going upward)

$$\begin{gathered} 20s=1600 \\ s=\frac{1600}{20} \\ s=80m \end{gathered}$$

Hence, the maximum height is $80m$.

Step 3: Calculation of net displacement and total distance covered

Total distance covered = $2\times$maximum height

$$\begin{gathered} =2\times 80m \\ =160m \end{gathered}$$

Net displacement is zero because initial position and final position are same.