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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)

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Solution

## (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, $20=0+\frac{1}{2}×g\mathit{×}{t}^{\mathit{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathit{⇒}{t}^{\mathit{2}}=\frac{40}{\mathrm{g}}=\frac{40}{10}=4\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathit{⇒}t=4\mathrm{s}$ (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, $v=0+10×4=40\mathrm{m}/\mathrm{s}$

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