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a freely falling object travells 9/25 of its total height in the last second then find the time of its fall

a freely falling object travells 9/25 of its total height in the last second then find the time of its fall

Grade:11

1 Answers

Arun
25750 Points
5 years ago
 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. ANS.

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