Question

A stone is dropped from a height of 20 m.
(a) How long will it take to reach the ground?
(b) What will be its speed when it hits the ground? (g = 10 m/s²)

Answer 1

(a) Distance travelled by the stone while falling from the height 20 m to the ground = 20 m

The initial velocity of the stone = 0

Acceleration of the stone = g

By using the second equation of motion, we can write, $s=ut+\frac{1}{2}g{t}^2$

Here, u = 0, s = 20 m, and 't' is the time taken by the stone to reach the ground.

So,

$$\begin{gathered} 20=0+\frac{1}{2}\times g\mathit{\times }{t}^{\mathit{2}} \\ \mathit{\Rightarrow }{t}^{\mathit{2}}=\frac{40}{\mathrm{g}}=\frac{40}{10}=4 \\ \mathit{\Rightarrow }t=4\mathrm{s} \end{gathered}$$

(b) Using the first equation of motion, we can write, $v=u+gt$ (Where 'v' is the speed of the stone when it hits the ground)

So, $v=0+10\times 4=40\mathrm{m}/\mathrm{s}$