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

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.

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