Question

a body of mass 2kg is dropped from a height of 1m .The kinetic energy as it touches the ground is

Answer 1

k.e= 1/2 mv²
but to find v we use
v²-u² =2as
v²-0= 2×10×1
then v =√20
now.K.E=1/2×2 ×√20² = 20 joules answer

Answer 2

Answer:

The kinetic energy as it touches the ground is 20J.

Explanation:

The body being dropped from a height acquires kinetic energy during the motion which  is calculated using equation of motion and formula of kinetic energy as follows -

Given: Initial velocity = u = 0

Acceleration a = g

Height h = 1 m

Final velocity v = ?

From equation of motion, we have –

$\bold{v^{2}-u^{2}=2 a s}$

$\begin{array}{l}{\mathrm{v}^{2}-\mathrm{u}^{2}=2 \mathrm{g} \times \mathrm{h}} \\ {\mathrm{v}^{2}-(0)^{2}=2 \times 10 \times 1} \\ {\mathrm{v}^{2}=20}\end{array}$

from the kinetic energy formula we have –

$\bold{K . E .=\frac{1}{2} m v^{2}}$

$\mathrm{K.E.}=\frac{1}{2} \times 2 \times 20$

K.E. = 20 J