Question

a construction worker drops a wrench from the top of an 80 m high tower. what is the velocity of the wrench when it strikes the ground

Answer 1

\(S=\frac{gt^2}2\)
\(v=gt\)
\(t=\sqrt{\frac{2S}{g}}\)
\(v=g\cdot\sqrt{\frac{2S}{g}}=\sqrt{2gS}\)

Answer 2

Do you know how to solve it

Answer 3

It is a solution, S - is a distance, v - velocity, g - acceleration = 9.8 m/s^2