Question

a stone is dipped from a height of 20 m what will be its speed when its ground g 10m/s​

Answer 1

solution

$s = ut + \frac{1}{2} at {}^{2} \\ = but \: \: u = 0 \\ so..........s = \frac{1}{2} at {}^{2}$

now...the speed is ...

$= at = 10 \times (2) \: \: m.s {}^{ - 1} = 20 \: m.s {}^{ - 1}$

$20 = \frac{1}{2} \times 10 \times t {}^{2} \\ = > t = \sqrt{4} = 2 \: \: sec$

Answer 2

Answer:

For a freely falling body

v^2 = u^2 + 2gh

v^2 = (0)^2 + 2(10)(20)

v^2 = 400

v = √400

v= 20m/s

The speed of a stone when it hits ground is 20m/s.

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