Question

A ball of mass 200 grams falls from hight of 5 metres what is it's kinetic energy when it just reached the ground

Answer 1

Kinetic energy =
$\frac{1}{2} m {v}^{2}$
m= mass = 200gm = 200/1000kg

= 1/5kg

but velovity become 0

therefore KE=0

but while reaching the ground velocity =
v^2-u^2=2as
v^2 =2×10×5 - u^2
(u=0)
v^2= 100-0
v=root 100
v=10.

then KE= 1/2×1/5×10×10
= 1/10×10×10
=10

KE = 10joule.

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Answer 2

mass = 200g => 0.2 kg

height = 5m

k.e = $\frac{1}{2} mv^{2}$

note: $v^{2} =u^{2} +2as$

here distance is height so s=h

K.E = $\frac{1}{2} (0.2) [2(9.8)(5)]$

initial velocity u = 0

and a = g= 9.8

k.E = 0.1(2)(9.8)(5)

K.E = 9.8