A friend offered me a brain teaser to which the solution involves a $195$ pound man juggling two $3$-pound balls to traverse a bridge having a maximum capacity of only $200$ pounds. He explained that since the man only ever holds one $3$-pound object at a time, the maximum combined weight at any given moment is only $195 + 3=198$ pounds, and the bridge would hold.
I corrected him by explaining that the acts of throwing up and catching the ball temporarily make you 'heavier' (an additional force is exerted by the ball to me and by me onto the bridge due to the change in momentum when throwing up or catching the ball), but admitted that gentle tosses/catches (less acceleration) might offer a situation in which the force on the bridge never reaches the combined weight of the man and both balls.
Can the bridge withstand the man and his balls?
Answer
Suppose you throw the ball upwards at some speed $v$. Then the time it spends in the air is simply:
$$ t_{\text{air}} = 2 \frac{v}{g} $$
where $g$ is the acceleration due to gravity. When you catch the ball you have it in your hand for a time $t_{\text{hand}}$ and during this time you have to apply enough acceleration to it to slow the ball from it's descent velocity of $v$ downwards and throw it back up with a velocity $v$ upwards:
$$ t_{\text{hand}} = 2 \frac{v}{a - g} $$
Note that I've written the acceleration as $a - g$ because you have to apply at least an acceleration of $g$ to stop the ball accelerating downwards. The acceleration $a$ you have to apply is $g$ plus the extra acceleration to accelerate the ball upwards.
You want the time in the hand to be as long as possible so you can use as little acceleration as possible. However $t_{\text{hand}}$ can't be greater than $t_{\text{air}}$ otherwise there would be some time during which you were holding both balls. If you want to make sure you are only ever holding one ball at a time the best you can do is make $t_{\text{hand}}$ = $t_{\text{air}}$. If we substitute the expressions for $t_{\text{hand}}$ and $t_{\text{air}}$ from above and set them equal we get:
$$ 2 \frac{v}{g} = 2 \frac{v}{a - g} $$
which simplifies to:
$$ a = 2g $$
So while you are holding one 3kg ball you are applying an acceleration of $2g$ to it, and therefore the force you're applying to the ball is $2 \times 3 = 6$ kg.
In other words the force on the bridge when you're juggling the two balls (with the minimum possible force) is exactly the same as if you just walked across the bridge holding the two balls, and you're likely to get wet!
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