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

Time of ascent =ug=4010 s=4 sfrom 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=10 m