Question

A stone is released from the top of a tower of height $19.6m$. Calculate its final velocity just before touching the ground.

Answer 1

Step 1: Given:

Height of the tower = $19.6m$

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

Step 2: Formula and calculation:

Using the third equation of motion

$$\begin{gathered} v^2=u^2+2gs \\ {v}^2=0+2\times 9.8\times 19.6 \\ {v}^2=19.6\times 19.6 \\ v=19.6m{s}^{-1} \end{gathered}$$

Hence, the speed of stone just before touching the ground is $19.6m{s}^{-1}$