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a body is dropped from a certain height such that the distance travelled in last second is equal to the distancetravelled in first three second.find its height
Asked by ritukkr4444 | 04 Jul, 2019, 07:06: PM
Distance travelled in first 3 seconds = (1/2)gt2 = (1/2)×9.8×9 = 44.1 m ..................(1)

Let the body takes n second to reach ground.

Velocity after (n-1) seconds = gt = 9.8×(n-1)

Distance Sn travelled at last nth second  is obtained from the formula , Sn = ut+(1/2)gt2
where u = initial velocity = 9.8×(n-1) and t = 1 s

Hence Sn = 9.8×(n-1) + (1/2)×9.8×1 = 9.8×(n-1) + 4.9 ....................(2)
It is given that both the distances given by eqn.(1) and (2) are equal.

Hence 9.8×(n-1) + 4.9 = 44.1  or  n = 5 s

Hence total height = (1/2)gt2 = (1/2)×9.8×52 = 122.5 m
Answered by Thiyagarajan K | 04 Jul, 2019, 07:51: PM
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