Find the final velocity of the body, if kinetic energy at ground=kinetic energy+potential energy.

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Solution

$\begin{matrix} \frac{1}{2} m u^{2} = \frac{1}{2} m v^{2} + m g h \frac{1}{2} \times 2 \times {(6)}^{2} = \frac{1}{2} \times 2 \times v^{2} + 2 \times 10 \times 1 36 = v^{2} + 20 v^{2} = 36 - 20 v^{2} = 16 v = \sqrt{16} = 4 \end{matrix}$

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Find the final velocity of the body, if kinetic energy at ground=kinetic energy+potential energy.

12mu2=12mv2+mgh12×2×(6)2=12×2×v2+2×10×136=v2+20v2=36−20v2=16v=√16=4

$\begin{matrix} \frac{1}{2} m u^{2} = \frac{1}{2} m v^{2} + m g h \frac{1}{2} \times 2 \times {(6)}^{2} = \frac{1}{2} \times 2 \times v^{2} + 2 \times 10 \times 1 36 = v^{2} + 20 v^{2} = 36 - 20 v^{2} = 16 v = \sqrt{16} = 4 \end{matrix}$