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Maximum height means v = 0. Then using v = vi + at, and a vertical axis with origin at the top of the building pointing upward, withv = 0, a = -9.8 m/s2 and vi = 14 m/s, the time to reach maximum height is t = (14 m/s)/(9.8 s) = 1.43 s
Now, using y = yi + vit + ½ at2 with the same vertical axis already mentioned, yi = 0, vi = +14 m/s, a = -9.8 m/s2, we get
ymax = (14 m/s)·(1.43s) - (4.9m/s2)(1.43 s)2 = 10 m
b) Keeping the same axis as before, and using vi = +14 m/s, a = -9.8 m/s2 and t = 4.5 s in y = yi + vit + ½ at2, y = (14 m/s)·4.5 s - 4.9m/s2(4.5 s)2 = -36.2 m. The result is negative as expected since the ground is in the negative side of the axis. The height of the building is then 36.2 m.
c) The velocity reaching the ground is v = 14 m/s - 9.8 m/s2·4.5 s = -30.1 m/s.
Regards
Arun(askIITians forum expert)
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