Posted by nj on Tuesday, September 16, 2008 at 10:17pm.
For constant acceleration with 0 initial velocity, the distance is:
distance = (1/2) (acceleration)(time^2)
If D= the total distance to the ground:
D = (1/2)g(TotalTime^2)
(1/2)D=(1/2)g(MidpointTime^2)
or
(1/2)g(TotalTime^2)= 2[(1/2)g(MidpointTime^2)]
(TotalTime^2) = 2(MidpointTime^2)
the problem states MidpointTime=1.6s
Substitute and solve for the total time
d=1/2at^2
a is the acceleration of gravity, t is given as 1.6s. Plug those in and solve for d. The problem states that is halfway.
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