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A clown starts to juggle balls. For one act, he throws a ball vertically upwards and runs to a cage 6m away. He runs at a constant speed of 3 m/s...

Hello!


It is obvious that the time needed for a clown to return is `t_0=4s` (twice 6m divided by 3m/s). A ball must return to the same height after the same time.


Choose a vertical upward axis starting at the initial height of a ball. Then the height of a ball is `H(t)=V_0 t-(g t^2)/2,` where `V_0` is the initial speed we have to find, `t` is time and `g` is the gravity acceleration.



We need `H(t_0)=0` (a ball returns at the same time a clown returns). It is a simple equation for `V_0,` `V_(0) t_(0)=(g t_0^2)/2,` or `V_0=(g t_0)/2.` In numbers it is `(9.8*4)/2=19.6 (m/s).` This is the answer for (i).



For (ii), use that speed of a ball is `V(t)=V_0 - g t,` and that at the maximum height the speed is zero. So the time when the maximum height is reached is `t_1=V_0/g,` and the corresponding height is `H(t_1)=(V_0^2)/(2g)=(g t_0^2)/8=(9.8*4^2)/8=19.6 (m).`


This is the answer for (ii).

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