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# a body dropped from top of the tower fall through 40m during the last two seconds of its fall the height of the tower

Asked by frozen.queen2488 1st June 2022, 6:23 PM
Let A be the point at Top of tower of height h .  Let C be the point at bottom of tower.

Let B be the point where the body reaches 2 second before reaching the bottom of tower .

At the point B , velocity v of body  is given as

v = g × (t-2 )   .............................. (1)

From B to C , the body has travelled 40 m distance in 2 seconds. hence we have

40 = [ g × (t-2) ] × 2 + (1/2) g 22  ...........................(2)

( equation of motion " S = u t + (1/2) g t2 " is used to get above equation (2)  )

By simplifying eqn.(2) , we get

2 g (t-1)= 40

Hence time t is calculated as

t = [ 40 / ( 2 × 9.8) ] +1 = 3.04 s

Total height of the tower is calculated as

h = (1/2) g t2 = (1/2) × 9.8 × 3.04 × 3.04 = 45.31 m
Answered by Expert 1st June 2022, 8:46 PM
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