Question

A ball is thrown downward from the top of a building 122 m tall with an initial speed of 38.0 m/s. What will be the speed of this ball when it reaches the ground?

Answer 1

`\begin{gathered} v_o=38.0\text{ m/s} \\ h=122\text{ m} \\ a=9.81m/s^2 \\ v_f=? \\ vf^2=vo^2+2ah \\ vf=√(vo^2+2ah) \\ vf=\sqrt{(38.0\text{ m/s})^2+2(9.81m/s^2)(122m)} \\ vf=\sqrt[]{1,444.0m^2/s^2+2,393.64\text{ }m^2/s^2} \\ vf=\sqrt[]{3,837.64m^2/s^2} \\ vf=61.95\text{ m/s} \\ \text{The s}peed\text{ is }61.95\text{ m/s} \end{gathered}`