If height is equal to 40m then its position after 1second from ground in a freely falling body?
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Let h = the height of the tower that needs to be determined.
Let t be the time of fall.
Then (t - 2) would be the time to reach the top of the 40 meter mark, and let d be the distance fallen till it reaches the 40 m mark.
Using kinematics, we can write: d = 1/2 g (t-2)^2
And H = 1/2 g t^2
H also = 40 + d
Then: H = 40 + 1/2 g (t-2)^2 = 1/2 g t^2
Expand: 40 +5 (t^2 - 4 t + 4) = 5 t^2
40 + 5 t^2 - 20 t + 20 = 5 t^2
Add like terms: - 20 t = - 60
t = 3 s
(t-2) = 1 second
In one second an object falls 5 m
Then H = 45 m
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