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A juggler is maintaining four balls in motion, making each in turn to rise to height of 20 m.Where will other three balls be at the instant, When one is just leaving his hand? $g= 10ms^{-2}$

$\begin{array}{1 1}(A)\;30m\\(B)\;20m \\(C)\;9 m \\(D)\;6m \end{array} $

1 Answer

The time of rise is u/g.
Here $u= 20 m/s^{-1}$ and $g= 10 ms^{-2}$
so a ball takes $\large\frac{20}{10}$
i.e 2s to rise.
It also takes 2s to fall . The total time for which the ball remains in air is 4s.
Since the juggler is maintaining 4 balls, he must be throwing them at a regular interval of 1 s. The position of the ball which in the air for $0,1,2,3s$ are as follows,
$s_0 = 1m \uparrow$
$s_1=20(1) -\large\frac{1}{2}$$(10)(1^2) =15 m \downarrow$
$s_2=20(2) -\large\frac{1}{2}$$(10)(2^2) =20 m \downarrow$
$s_3=20(3) -\large\frac{1}{2}$$(10)(3^2) =15 m \downarrow$
When a fresh ball is projected , the two balls are at 15 m height travelling in opposite direction and one ball is at the highest level of 20 m
Hence B is the correct answer.
answered Jul 25, 2014 by meena.p

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