$\begin{array}{1 1}(A)\;\frac{1}{100} \\(B)\;10 \sqrt {10} ms^{-1} \\(C)\;\frac{1}{300}\\(D)\;2 \end{array} $

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Comparing $y=x -\large\frac{x^2}{100}$

with $y=\tan\theta x-\large\frac{1}{2} \frac{g}{u^2 \cos ^2 \theta}$$x^2$

we get that the value of $\theta=45$

the velecity of projection is given by

$v_0^2 =\large\frac{100 g}{2 \cos ^2 \theta_0} =\frac{100(10)}{2(1/ \sqrt 2 )^2}$$=1000(m/s^2)$

$\qquad= v_0 = 10 \sqrt {10} ms^{-1}$

Hence B is the correct answer.

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