Kinematics Problem

Here, we have the following values,

$Height=s=20m$

$Acceleration=g=10m/s^{2}$

$Initial \quad velocity=0m/s$

$Final \quad velocity=?$

Substituting the values in the third equation of motion, i.e. $v^{2}-u^{2}=2gs$

We get, $v^{2}=2×10×20$

$v=\sqrt{400}$

Therfore, $v=\boxed{20m/s}$

K.E = P.E

$\frac{1}{2} m v^2 = m g h$

$\frac{1}{2} v^2 = g h = 200$

$v = 20 m/s$

distance(s) = 20m initial velocity(u) = 0m final velocity = v acceleration = 10m/s v^2 - u^2 = 2as v^2 = 2 * 10 * 20 v^2 = 400 v = 20. Thus, final velocity is 20 m.

V^2 = 2as

V^2 = 400

V=20 m/s

Very simple just use laws of motion Apply the formulas carefully Here we apply 3rd law of motion V^2-u^2=2ag