A stone dropped from the top of a tower is found to travel 5/9 of the ...
Explanation:
Let's assume the height of the tower is 'h' and time of fall is 't'.
Given, during the last second of its fall, the stone covers 5/9th of the height of the tower.
So, the distance covered in the last second = 5/9 * h
We know that the distance covered by a freely falling body in t seconds is given by the formula:
S = ut + (1/2)gt^2
Where, u = initial velocity = 0 m/s (as the stone is dropped)
g = acceleration due to gravity = 9.8 m/s^2 (downwards)
So, the distance covered by the stone in t seconds = h
Therefore, we can write the equation as:
h = 0 + (1/2)gt^2
=> t^2 = 2h/g
=> t = √(2h/g)
Now, we need to find the value of 't'.
So, let's substitute the given value of distance covered in the last second:
5/9h = u(1) + (1/2)g(1)^2
=> 5/9h = 0 + (1/2)g
=> (5/9)h = (1/2)g
Substituting the value of 'g' in terms of 'h':
(5/9)h = (1/2)(2h/t^2)
=> t^2 = (9/5) * 2
=> t = √(18/5) seconds
=> t = 3 seconds
Therefore, the time of fall is 3 seconds.
A stone dropped from the top of a tower is found to travel 5/9 of the ...
Let the stone travels for n second. according to question, distance in nth second =5/9 of height of tower so from the data, u+n^2*a/2= total height u+(2n-1) =5/9*total height as initially stone at rest, u=0 by dividing the equation and solving it, you get n=3 second
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