By the definition, power is work performed at a time unit (work divided by time). Therefore in this problem the total work done is `P*T,` where `P` is the given power and `T` is the given time.

Also it is known that the amount of work performed on a body...

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By the definition, power is work performed at a time unit (work divided by time). Therefore in this problem the total work done is `P*T,` where `P` is the given power and `T` is the given time.

Also it is known that the amount of work performed on a body adds to its total energy. The total energy of the sacks consists of the kinetic energy `(m V^2)/2` and the potential energy `mgh.` So `(mV_e^2)/2 + mgH = PT` because the initial speed is probably zero. Here `H` is the height in question and `V_e` is the end speed.

To maximize `H` we need to minimize `V_e^2,` and its minimum is zero. In this case `mgH = PT` and `H = (P T)/(mg).`

Two sacks have total mass `m` of `100 kg.` Thus the numerical answer is about `(480*2)/(100*9.8) approx 1.0 m.`

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