What was the total time of fall?

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The discussion revolves around calculating the height from which a ball was dropped and the total time of its fall, given that it covers three-sevenths of the distance in the last two seconds. Participants clarify that the final velocity is not needed to determine the height, as the height can be expressed in terms of time and acceleration due to gravity. The equation derived from the problem is s(T- 2) = h - 4.905(T - 2)^2 = (3/7)h, which can be solved for height. One user attempted to calculate the height and arrived at 45.78 meters but found the result unsatisfactory. The conversation emphasizes the importance of correctly applying the kinematic equations to solve the problem.
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A ball, dropped from rest, covers three-sevenths of the distance to the ground in the last two seconds of its fall.

(a) From what height was the ball dropped?
_m
(b) What was the total time of fall?
_s

For a, don't i need the final velocity to solve the problem? Vi=o A=-9.81?
2 days ago - 1 day left to answer.
 
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keweezz said:
A ball, dropped from rest, covers three-sevenths of the distance to the ground in the last two seconds of its fall.

(a) From what height was the ball dropped?
_m
(b) What was the total time of fall?
_s

For a, don't i need the final velocity to solve the problem? Vi=o A=-9.81?
2 days ago - 1 day left to answer.
No, you don't need the final velocity (although you can calculate it after you have solved b). I presume you know that the height of the ball after t seconds is s= h- (1/2)At2= h- 4.905t2. It will hit the ground when s= h- 4.905t2= 0 so the time it hits the ground, T, is T= \sqrt{h/4.905}.
the ball "covers three-sevenths of the distance to the ground in the last two seconds of its fall." which tells us that s(T- 2)= h- 4.905(\sqrt{h/4.905}- 2)^2= (3/7) h.
Solve that equation for h.
 


HallsofIvy said:
No, you don't need the final velocity (although you can calculate it after you have solved b). I presume you know that the height of the ball after t seconds is s= h- (1/2)At2= h- 4.905t2. It will hit the ground when s= h- 4.905t2= 0 so the time it hits the ground, T, is T= \sqrt{h/4.905}.
the ball "covers three-sevenths of the distance to the ground in the last two seconds of its fall." which tells us that s(T- 2)= h- 4.905(\sqrt{h/4.905}- 2)^2= (3/7) h.
Solve that equation for h.



s(T- 2)= h- 4.905(\sqrt{h/4.905}- 2)^2= (3/7) h. For that part, i just do h- 4.905(\sqrt{h/4.905}- 2)^2= (3/7) h that and solve for h?

i got 45.78 for h, but it didn't work out ;(
 
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