here's the solution : -
we know, kinetic energy (k.e) =
[tex] \dfrac{1}{2} mv { }^{2} [/tex]
where,
So,
=》
[tex]100 = \dfrac{1}{2} \times 0.5 \times {v}^{2} [/tex]
=》
[tex] \dfrac{100 \times 2}{0.5} = {v}^{2} [/tex]
=》
[tex] \dfrac{2000}{5} = {v}^{2} [/tex]
=》
[tex]400 = {v}^{2} [/tex]
=》
[tex]v = \sqrt{400} [/tex]
=》
[tex]v = 20\: \: ms {}^{ - 1} [/tex]
