question_answer

A body of mass 1 kg is allowed to fall freely under gravity. Find the K.E. of the body, 5 sec after it starts falling. (Take\[\text{g}=\text{1}0\text{m}{{\text{s}}^{-\text{2}}}\])

Answer:

Given,   \[m=1\,\,kgt=5s\]                 \[g=10m/{{s}^{2}}u=0\]                 \[K.E.=?\] We know, \[K.E.=\frac{1}{2}m{{v}^{2}}\]                              ???.(1) For this first we have to find ?v? From the relation \[S=ut+\frac{1}{2}g{{t}^{2}}\] we get, \[S=0\times 5+\frac{1}{2}\times 10\times {{(5)}^{2}}=125\,m\] So, we get displacement value now we have to find ?v? value. From the relation, \[{{v}^{2}}-{{u}^{2}}=2gS\] we get, \[{{v}^{2}}-{{0}^{2}}=2\times 10\times 125\] \[\Rightarrow {{v}^{2}}=2500\] Applying the above value in equation (1) \[\because K.E.=\frac{1}{2}\times 4\times 2500=5000J\]