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nasticia
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
\(S=\frac{gt^2}2\) \(v=gt\) \(t=\sqrt{\frac{2S}{g}}\) \(v=g\cdot\sqrt{\frac{2S}{g}}=\sqrt{2gS}\)
Do you know how to solve it
It is a solution, S - is a distance, v - velocity, g - acceleration = 9.8 m/s^2