$\begin{array}{1 1}(A)\;10 \sqrt 7 \\(B)\;6 \sqrt 2 \\(C)\;5 \sqrt 5 \\(D)\;6.7 \sqrt 2 \end{array} $

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After 4 seconds , the particle returns to the same mark.

If its velocity when crossing 15 cm high mark is v then in 4s its displacement is zero.

Therefore $0= v \times 4 -\large\frac{1}{2}$$ \times 10 \times 4^2$

$v= 20 ms^{-1}$

If the initial velocity of projection is u, then

$v^2=u^2 +2as $ gives

$20^2 =u^2 -2(10)(5)$

$u^2= 700$

$u= 10 \sqrt 7 ms^{-1}$

Hence A is the correct answer.

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