Answer:
6.26 m/s
Explanation:
From the given information:
Using the conservation of Energy;
M.E at A =M.E at B
[tex]\dfrac{1}{2}mv_A^2 + mgh _A = \dfrac{1}{2}mv_B^2 + mgh_B[/tex]
Replacing the values given in the equation:
[tex](1) (9.8) (2 \ m) +\dfrac{1}{2}(0)^2 = 1 \times 9.8 \times 0 + \dfrac{1}{2}(1) \times v_B^2 \\\\ 19.6 + 0 = 0 + 0.5v_B^2\\ \\ v_B^2 = \dfrac{19.6}{0.5} \\ \\ v_B^2 = 39,2 \\ \\ v_B = \sqrt{39.2} \\ \\ \mathbf{v_B = 6.26 \ m/s}[/tex]
