Question

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

(take $g=9.8m/s^2$)

Answer 1

Step 1: Given data and the diagram of the situation

Height of tower (h) = $19.6m$

Acceleration due to gravity = a= $g=+9.8m/s^2$.

Note: Here the stone is going along with the gravity, so the value of g is taken to be as positive.

Final velocity (v) = ?

The diagrammatic representation of the given case:

Step 2: Formula used

According to $3^{rd}$ equation of motion

$v^2=u^2+2as.......\left(a\right)(s=h=displacement)$

Step 3: Finding the final velocity of the stone

We know that, there is no velocity required to drop a stone from height. i.e u=0

On, putting all the given values in equation (a), we get

$$\begin{gathered} v^2=0+2\left(9.8m/s^2\right)\times 19.6m \\ {v}^2=19.6\times 19.6 \\ v=\sqrt{19.6\times 19.6} \\ v=19.6m/s \end{gathered}$$

Hence, the final velocity attained by the stone is $19.6m/s$.