Let the bucket be the mass 1 and Indy be the mass 2. See the figure attached for the forces acting on both masses.

For the bucket, 1 we can write:

`F = G_1 +m_1*a`

For Indy, 2 we can write

`F +m_2*a =G_2`

Thus

`G_1 +m_1*a +m_2*a =G_2`

`a(m_1+m_2) =g(m_2-m_1)`

`a = g*(m_2-m_1)/(m_1+m_2)` (1)

Now for the uniform accelerated motion we have the equation relating the space to the time

`(s =30 ft =30*0.305 =9.15 m)`

`s = (a*t^2)/2 => a= 2*s/t^2 =2*9.15/4^2 =1.14 m/s^2`

Now from expression (1) we obtain

`(m_2-m_1)/(m_1+m_2) =1.14/9.81`

`1.14*m_1 +1.14*m_2 =9.81*m_2 -9.81*m_1`

`10.95*m_1 =8.67*m_2`

`m_1 =8.67/10.95 *m_2 =0.792*180 =142.52 lb =64.633 kg`

**The total mass of the bucket and its extra weights need to be 64.63 kg**

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