question_answer

A juggler maintains four balls in motion, making each of them to rise a height of 20 m from his hand. What time interval should he maintain, for the proper distance between them? \[\left( \text{g}=\text{1}0\text{ m}/{{\text{s}}^{\text{2}}} \right)\]            

Answer:

                Here, S = 20 m. Using; \[{{\upsilon }^{\text{2}}}={{\text{u}}^{\text{2}}}+\text{2as}\]. For upward motion, we have \[0={{\text{u}}^{\text{2}}}+\text{2}\times \left( -\text{1}0 \right)\times \text{2}0\]or \[\text{u}=\text{ 2}0\text{ m}/\text{s}\]. If t is the total time of flight of the ball in going up and coming back, then total displacement in time t is zero i.e, S = 0. Using the relation, \[\text{S}=\text{ut}+\frac{1}{2}\text{a}{{\text{t}}^{\text{2}}}\], we have \[0\text{ }=\text{ 2}0\text{t }+\frac{1}{2}~\left( -\text{1}0 \right){{\text{t}}^{\text{2}}}=\text{ 2}0\text{t }-\text{5}{{\text{t}}^{\text{2}}}\] or r = 4sec. time interval of each ball = \[\frac{4}{4}\] = 1 sec.