Question

A stone is dropped from a height of 20 m.
1. How long will it take to reach the ground? (ii) What will be its speed when it hits the ground ? ( g = 10 m/s-²)​

Answer 1

Explanation:

S=ut+1/2at^2

u=0,a=10,s=20(given)

t=2sec

v=u+at

u=0,a=10,t=2

v=20m/s

Answer 2

$\huge \purple \star \purple {Answer} \huge\purple \star$

$height. \: s = 20m \\ initial \: velocity. \: u = 0 \\ acceleration \: due \: to \: gravity. \: g \: = \frac{10m}{ {s}^{2} } \\ final \: velocity \: = \: x \\ time \: taken \: = x \\ 1) \: \: using \: relation \: \\ \\ s \: = ut \: + \frac{1}{2} {gt}^{2} \\ 20 = 0 \times t + \frac{1}{2} \times 10 \times {t}^{2} \\ 20 = 0 + {5t}^{2} \\ {t}^{2} = \frac{20}{5} = 4 \\ t = \sqrt{4} = 2s \\ \\ for \: a \: freely \: falling \: body \: = \\ {v}^{2} = {u}^{2} + 2gh \\ {(0)}^{2} + 2 \times( 10) \times( 20) \\ so \: {v}^{2} = 400 \\ v = \sqrt{400} = 20 \frac{m}{s}$