A large heavy box is sliding without friction down a smooth plane...

u is the relative velocity of the particle with respect to the box. Resolve u.
& ${u}_{x}$ is the relative velocity of the particle with respect to the box in x& –& direction.

& ${u}_{y}$ is the relative velocity of the particle with respect to the box in y –& direction.

Since there is no velocity of the box in the y& –& direction, therefore this is the vertical velocity of the particle with respect to ground also.

Y –& direction motion (Taking relative terms w.r.t. box)

& ${u}_{y} = + usin\alpha$

${a}_{y} = - gcos\theta$

${S}_{y} = 0$ (activity is taken till the time the particle comes back to the box)

& ${t}_{y} = t$

Using, $S = ut + \dfrac{1}{2} at^{2}$ we get–

$$\begin{array}{ll}0 = ( usin\alpha )t - \dfrac{1}{2} gcos\theta \times t^{2} \\ ⇒ t = 0ort = \dfrac{2usin\alpha}{gcos\theta}\end{array}$$

X-direction motion (Taking relative terms w.r.t. box)

$$\begin{array}{ll}{u}_{x} = + ucos\alpha \\ {a}_{x} = 0\end{array}$$

Again, using $S = ut + \dfrac{1}{2} at^{2}$ we get –

& $⇒$ ${S}_{x} = ucos\alpha \times \dfrac{2usin\alpha}{gcos\theta} = \dfrac{u^{2} \sin 2\alpha}{gcos\theta}[ ∵ {t}_{x} = \dfrac{2ucos\alpha}{gcos\theta} ]$