A body is dropped from a certain height. 02 seconds later another body is projected vertically upward from the same point at 20m/s. The distance between the first and second bodies 5 seconds after the projection of the second body is.........m.
(acceleration due to gravity = 10 m/s2)
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Explanation:
- dropped body u=0
- s=ut+1/2at²
- s=0+1/2(10)(5)²
- s=125m
- projected body u=20
- s at 5 seconds
- s=20(5)-10(25)
- s=-150
- so difference between their distance is 125-(-150)
- =275
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