Answer:
[tex]v_{yf}=6.26m/s[/tex]
Explanation:
We have to focus only on the vertical direction: the puck falls 2m with a null vertical initial velocity. We use the equation:
[tex]v_f^2=v_i^2+2ad[/tex]
which for the vertical direction is:
[tex]v_{yf}^2=v_{yi}^2+2a_y\Delta y[/tex]
or:
[tex]v_{yf}=\sqrt{v_{yi}^2+2a_y\Delta y}[/tex]
Taking the downward direction as positive (which means a positive gravitational acceleration and a positive displaceent) we have for our values:
[tex]v_{yf}=\sqrt{(0m/s)^2+2(9.8m/s^2)(2m)}=6.26m/s[/tex]
