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Answer to Question #78324 in Physics for Katie
Question #78324
A box is dropped from an 8.45 m bridge
Calculate the speed of the box just before it hits the ground.
Calculate the speed of the box just before it hits the ground.
Expert's answer
From the conservation of energy:
(mv^2)/2=mgh
v^2/2=gh
The speed of the box just before it hits the ground:
v=√2gh=√(2(9.8)(8.45))=12.9 m/s.
(mv^2)/2=mgh
v^2/2=gh
The speed of the box just before it hits the ground:
v=√2gh=√(2(9.8)(8.45))=12.9 m/s.
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