Question

A rock is thrown downward from the top of a 40.0m tall tower with an initial speed of 12 m/s. Assuming negligible air resistance, what is the speed of the rock just before hitting the ground?

Answer 1

$h=\frac{V^2-V_0^2}{2g};\\V=\sqrt{2gh+V_0}=\sqrt{2\times9.8\times40+12}=28.21\frac{m}{s};\\Answer: the \;speed\ of\ the\ rock\ is \;28.21\frac{m}{s};$