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A man standing on top of a building throws a ball vertically upwards with a velocity of 14m/s. The ball reaches the ground 4.6s later.FindMaximum height reached by the ballHow high is the buildingVelocity with which the ball strikes the ground( g=9.8m/s2)
Dear student 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 sNow, 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 getymax = (14 m/s)·(1.43s) - (4.9m/s2)(1.43 s)2 = 10 mb) 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.RegardsArun(askIITians forum expert)
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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