question_answer

A man drops a 10 kg rock from hte top of a 5m ladder. What is its K.E., when it reaches the ground? (Take \[\text{g}=\text{1}0\text{m}{{\text{s}}^{-\text{2}}}\])

Answer:

Given,   \[m=10\,kgh=5m\]                 \[g=10\,m/{{s}^{2}}K.E.=?\] We know, \[K.E.=\frac{1}{2}m{{v}^{2}}\] First we have to find, ?v? From the relation, \[{{v}^{2}}-{{u}^{2}}=2gh\] We get, \[{{v}^{2}}={{0}^{2}}=2\times 10\times 5(\because u=0)\] \[\Rightarrow {{v}^{2}}=100\] \[\therefore K.E.=\frac{1}{2}\times 10\times 100=500J\]