While riding an open freight elevator upwards at 5.4 m/s, a carpenter drops his hammer off the side. It hits the ground 2.4 s later. How far above the ground was the carpenter when he dropped the hammer?
Let H is the hight of elevator when the carpenter drop hammer.
Use kinematic equation,
-H = voy*t - 0.5*g*t^2
= 5.4*2.4 - 0.5*9.8*2.4^2
= -15.26
H = 15.26 m <<<<<<<--------AAnswer
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