A body dropped from top of a tower covers a distance 7x in last second of its journey where x is distance covrd in first second.?

After first second:

v = vo + gt

v = 0 -9.8(1)

v = -9.8m/s

v^2 = vo^2 + 2gΔy

(-9.8)^2 = 0 + 2(-9.8)Δy

96.04 = -19.6Δy

Δy = -4 9m in the first second.

After the final second, the body drops 7(-4.9) = -34.3m and its displacement is given by

Δy = vot + 1/2gt^2

0 - 34.3 = vo(1) + 1/2(-9.8)(1)^2

-34.3 = vo - 4.9

vo = -29.4m/s

We can now find the displacement between the final velocity after first second and initial velocity before final second:

v = vo + gt

-29.4 = -9.8 - 9.8t

-19.6 = -9.8t

t = 2s

So the total time is T = 1 + 2 + 1 = 4s