A particle is projected from a point on a fixed smooth inclined plane of angle θ=37^° (sin 37^°=0.6) with speed u in a direction perpendicular to the…

Resolve motion along and perpendicular to the plane. Perpendicular to the plane the acceleration is $g\cos\theta=8$ and the speed reverses at each elastic impact, so every flight lasts $T=2u/(g\cos\theta)=u/4$. Along the plane the acceleration is $g\sin\theta=6\ \mathrm{m\,s^{-2}}$ down the slope and the along-plane velocity is unchanged by the impact. Hence the along-plane velocity at the start of the $k$-th flight is $(k-1)\,g\sin\theta\,T$ and the distance covered in that flight is $d_k=\tfrac12 g\sin\theta\,T^2(2k-1)$. The distance between the 3rd and 4th impacts is the 4th flight: $d_4=\tfrac{7}{2}\cdot 6\cdot\frac{u^2}{16}=\frac{21u^2}{16}$. Setting this equal to $21$ gives $u^2=16$, so $u=4\ \mathrm{m\,s^{-1}}$. Check: with $u=4$, $T=1\ \mathrm{s}$ and $d_k=3,9,15,21\ \mathrm{m}$, so the 4th flight is $21\ \mathrm{m}$.