A force F = bt (where b is a constant) is applied at an angle to a mass m kept on a smooth horizontal plane. The velocity of mass m at the moment of its breaking off the plane is 

$\begin{array}{l}bt\cos \alpha =m\frac{dv}{dt} ......\left(i\right)\\ N+bt\sin \alpha =mg ........\left(ii\right)\end{array}$

For breaking off N=0

$\text{ Therefore from (ii), }t=\frac{mg}{b\sin \alpha }=t_0\text{ (say) }$

$\text{ From (i), }m{\int }_0^v dv=(b\cos \alpha ){\int }_0^{t_0} tdt$

$\begin{array}{l}mv=(b\cos \alpha )\frac{t_0^2}{2}=\frac{(b\cos \alpha )}{2}\frac{m^2{g}^2}{b^2{\sin }^2\alpha }\\ v=\frac{m{g}^2\cos \alpha }{2b{\sin }^2\alpha }\end{array}$