Your clue is impossible, water release at a velocity of 20m/s cannot lift a 10 kg garbage can under earth's gravity.
F=lambda 2v^2 (assuming elastic collision)
= 2mv^2/d
= 2(dm/dt)(v^2)/u
= 20N
which can lift a can of 20/9.81 kg
so your water can only lift a max of 2.04 kg garbage can.
An inverted garbage can of Weight W is suspended in air by water from a geyser. The water shoots up from the ground with a speed u, at a constant rate dm/dt. Find the maximum height at which the garbage can rides.What assumption must be fulfilled for the maximum height to be reached?
Clue: If u=20m/s, W=10kg, dm/dt=0.5 kg/s, then hmax= 17m.
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