For an observer on trolley, direction of projection of particle is shown in the figure, while for observer on ground ball rise vertically. The maximum height reached by ball from trolley is (take $g=10m/s^2$)

The correct option is B $15m$
Let $v$ be the magnitude of velocity of ball w.r.t ground and $v_{T_g}$ be the magnitude of velocity of trolley w.r.t ground.
From the ground frame of reference, horizontal velocity of ball w.r.t ground is zero.
We can write velocity of ball w.r.t trolley in component form as
$v_{b_T}=-v\cos {60}^{\circ }\hat{i}+v\sin {60}^{\circ }\hat{j}$
And from the equation of relative motion,
$v_{b_T}=v_{b_g}-v_{T_g}$
$\Rightarrow -v\cos {60}^{\circ }\hat{i}+v\sin {60}^{\circ }\hat{j}=v_{b_g}\hat{j}-10\hat{i}$
On Equating horizontal component (along x- axis), we get
$v\cos 60°=10$
$\Rightarrow v=20m/s$
We can calculate maximum height by using
$H=\frac{v^2{\sin }^2\theta }{2g}=\frac{{20}^2{\sin }^2{60}^{\circ }}{2g}=15m$