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Grade: 9


A body dropped from the top of a tower clears 9/25 of the total height of the tower in its last second of flight. What is the height of the tower?

2 years ago

Answers : (2)

24742 Points
Dear student
9/25 H = Ut + 1/2 gt^2; where t = 1 second and g is g. We need to find U, the initial speed as the body enters that last 9/25 of H the height of the tower. 

And, discounting air resistance as nothing's been given for that, U = sqrt(2g(16/25)H) assuming the drop means no initial speed at the top. Note 16/25 H is the height the body dropped up to the last second. 

So 9/25 H = sqrt(2g(16/25)H) + 4.9 and we define H = h^2 so we rewrite: 

9/25 h^2 = h (4/5) sqrt(2g) + 4.9 and the quad is 9/25 h^2 - (4/5) sqrt(2g) h - 4.9 = 0, which we solve for h = 11.068 
and H = h^2 = 122.5 m. 
2 years ago
na batau
17 Points

Ball travels 9/25 of the height (say x) in the t second(last second)

Hence it travelled 16/25 x in (t-1) seconds.

and the total distance in t seconds.

The ball is dropped hence initial velocity is zero and gravity is 9.8m/s^2 or g

Hence distance travelled is

s = 1/2 * a * t^2 = (1/2) g t^2

==> x = (1/2) gt^2 ______(1)

And 16x/25 = (1/2)g(t-1)^2 _______(2)

Divide the equations (2) by (1)

16/25 = (t-1)^2/t^2

==> 16t^2 = 25(t-1)^2 _____(3)

Which gives t = 5 seconds (Time cannot be negative, atleast not in this case)

Hence height x = (1/2)*9.8*(5^2) = 122.5 meters

or 125 meters (if you take g=10m/s^2)

2 years ago
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