$\begin{array}{1 1}(A)\; \frac{1}{3} \;m\\(B)\;\frac{2}{3}\;m \\(C)\;\frac{3}{4} \;m\\(D)\;\frac{4}{3}\;m \end{array} $

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Using the equation of trajectory and applying for the point $B( 2,h)$ and also for the point $C(6,0)$ we get,

$h= 2 \tan 45^{\circ}- \large\frac{g(2)^2}{2 v_0 ^2 \cos ^2 45^{\circ}}$------(i)

$0= 6 \tan 45^{\circ}$$ - \large\frac{g(6)^2}{2v_0^2\cos ^2 45^{\circ}}$-----(ii)

Putting the value of $\large\frac{g}{2v_90^2 \cos ^2 45^{\circ}}$ from (ii) and (i) we

$h= 2 - \large\frac{1}{6}$$(2)^2 =\large\frac{4}{3}$$\;m$

Hence D is the correct answer.

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