Question

A ball of mass 2 kg is kept on a tower of height 330 M find its potential energy at this point if it is not it if it is allowed to fall freely find its kinetic energy will it when it just touches the ground

Answer 1

m=2kg
h=330m
P.E.=mgh
=2×9.8(g=9.8m/s)×330
=6468J
K.E.=1/2 mv square
$v ^{2} = u { }^{2} + 2gh$
$v {}^{2} = 0 + 2 \times 9.8 \times 330$
$v { }^{2} = 6468$

kinetic energy is equals to half MV square we can put the value of V square =6468
KE is equals to 1 / 2 × 2 × 6468 we can cancel 2 by 2 and the kinetic energy is equal to 6 4 6 8 J.


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Answer 2

Answer:

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