Solving Kinmatics Problem - Ball Dropped from Rest

[tony873004]

A ball, dropped from rest, covers 2/7 of the distance to the ground in the last two seconds of its fall.

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

If I can figure out either part a or part b, the other part will be easy.

I have a feeling that the solution to this will involve simultaneous equations. But I don't really know where to begin aside from making a list of everything I know:

d_{t} = d_{1} + d_{2}
d_{t} = \frac {2}{7} d + \frac {5}{7} d

t_{t} = t_{1} + t_{2}
t_{t} = t_{1} + 2 seconds

v_{i_{t}} = 0
v_{f_{t}} = v_{f_{2}}

d_{1} = \frac{5}{7}d_{t}
t_{1} = t_{t} - t_{2}
t_{1} = t_{t} - 2 seconds

v_{i_{1}} = 0
v_{f_{1}} = v_{i_{2}}

d_{2} = \frac{2}{7} d_{t}
t_{2} = 2 seconds
v_{i_{2}} = v_{f_{1}}
v_{f_{2}} = v_{f_{t}}

Can anyone suggest a starting point?

 

[Pyrrhus]

Well, i will divide it in 2 parts, then i'll relate the equations with their link variable, the first part final speed, which will be the second part initial speed, Also i will use on the first part the equation without time, while on the second part i will use the displacement equation with time.

 

[maverick280857]

Suppose at any time the y coordinate of the ball is y(t). Then the distance it has covered equals (h-y(t)). So,

(h-y(t = T)) - (h-y(t = T-2)) = \frac{2h}{7}

Additionally, y(t = T) = 0.

Cheers
vivek