A particle is projected vertically upwards . It attains a height h after 2 seconds and again after 10 seconds . The speed of the particle at the height h is numerically equal to

$\begin{array}{1 1}(A)\;g \\(B)\;2g \\(C)\;4g \\(D)\;6g \end{array}$

Let the desired velocity be u.
As the particle returns to the same point in $(10-2)$ i.e 8 seconds , its displacement in 8 s is zero.
Using $s=ut+\large\frac{1}{2}$$at^2, we get, 0=u(8) -\large\frac{1}{2}$$g(8)^2$
$u=4g$
Hence C is the correct answer.