Question

A body of mass 10kg is dropped from a point where it possess on energy 100 j . then the KE when it reaches ground would be​

Answer 1

Answer:

change in P.E = change in K.E

$mgh \: = \: \frac{1}{2} mv^{2}$

mgh = 100 j

v = √ 2 P.E/m

= √ 200/10= √ 20 = 2√5 m/sec