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Vertical projection from ground

Question

A small body is projected vertically upward with a velocity of 40 m/s. Then the length of path traversed by the body in the time interval from t = 3 s to t = 5 s is 

Moderate
Solution

Time of ascent =ug=4010 s=4 s

from symmetry of vertical motion, we can write, distance covered in the last second of ascent = distance covered in the first second of descent. 

 length of path traversed =2×12g. t2=10m



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