In a projectile motion let $v_x$ and $v_y$ be the horizontal and vertical components of velocity at any time $t$ and $x$ and $y$ be the displacements along horizontal and vertical from the point of projection at any time $t$. Then

The correct options are
A $v_y-t$ graph is a straight line with negative slope and positive intercept.
B $x-t$ graph is a straight line passing through origin.
D $v_x-t$ graph is a straight line parallel to $t-$ axis.
If $u$ is the initial speed and $\theta$ is the angle of projection, then velocity of projectile along vertical direction is given by
$v_y=u\sin \theta -gt$
$\therefore$ $v_y-t$ graph is a straight line with negative slope and positive intercept.
Position of projectile at time $t$ along horizontal direction is given by
$x=\left(u\cos \theta \right)t$
$\therefore$ $x-t$ graph is a straight line passing through origin.
Position of projectile at any time $t$ along the vertical direction is given by
$y=\left(u\sin \theta \right)t+\frac{1}{2}g{t}^2$
$\therefore$ $y-t$ graph is a parabola.
Since projectile has uniform velocity along horizontal direction, we can say that
$v_x=u_x=u\cos \theta$
$\therefore$ $v_x-t$ graph should be a straight line parallel to $t-$ axis.
Thus, options (a), (b) and (d) are correct answers.