An object falls from rest from the top of a tower.The object takes 0.40s to fall from the mid-point of the tower to the ground.Calculate the time taken for the object to fall from the top of the tower to the ground.（Neglect air resistance）
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Physics question（Ans：1.37）求steps?
2016-11-27 5:36 am
回答 (1)
2016-11-28 8:23 am
Let 2L be the height of the tower.
Hence, using the equation of motion: s = ut + (1/2)at^2
From top of tower to ground: 2L = (1/2)g.(t + 0.4)^2 --------- (1)
(where g is the acceleration due to gravity, taken to be 10 m/s^2; and t is the time taken to fall from top to middle of tower).
From middle of tower to ground: L = (gt).(0.4) + (1/2)g(0.4)^2 -------- (2)
The term gt is the velocity of object at middle of tower.
(1)/(2): 2 = (g/2)(t+0.4^2)/[(gt(0.4) + (g/2)(0.4)^2]
After simplifying: 5t^2 - 4t - 0.8 = 0
solve for t gives t = 0.966 s (the -ve t value is rejected)
Therefore, the total time taken to fall from top of tower to ground
= (0.966 + 0.4) s = 1.366 s
Hence, using the equation of motion: s = ut + (1/2)at^2
From top of tower to ground: 2L = (1/2)g.(t + 0.4)^2 --------- (1)
(where g is the acceleration due to gravity, taken to be 10 m/s^2; and t is the time taken to fall from top to middle of tower).
From middle of tower to ground: L = (gt).(0.4) + (1/2)g(0.4)^2 -------- (2)
The term gt is the velocity of object at middle of tower.
(1)/(2): 2 = (g/2)(t+0.4^2)/[(gt(0.4) + (g/2)(0.4)^2]
After simplifying: 5t^2 - 4t - 0.8 = 0
solve for t gives t = 0.966 s (the -ve t value is rejected)
Therefore, the total time taken to fall from top of tower to ground
= (0.966 + 0.4) s = 1.366 s
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https://hk.answers.yahoo.com/question/index?qid=20161126213620AAvu77N