Question

a stone of mass 500g is dropped from a helicopter flying at a hight of 490m.what will be the kinetic energy of rhe object just before touching the ground​

Answer 1

${\huge {\mathfrak {Answer :-}}}$

Let, the velocity of the stone just before touching the ground be ' v '.

So, v^2 = u^2 + 2 g h.

Or, v^2 = 2 * 9.8 * 490.

Or, v^2 = 2 * 9.8 * 10 * 49

Or, v^2 = 98 * 98

Or, v = 98.

Now, kinetic energy will be

= 1/2 * 0.5 * (98)^2

= 49 * 49

= 2401 joules.

$<b>$ That's it..