One second from its launch, a projectile is moving at 45" to the horizontal, and one second later, it is moving horizontally. Its projection speed is (take $\left.g=10\mathrm{m}/{\mathrm{s}}^2\right)$:

Let the velocity of the projection be $\overrightarrow{u}=\left(u_x\right)\widehat{i}+\left(u_y\right)\widehat{j}$.

Then, one second later from the time of projection, its velocity wilt be ${\overrightarrow{v}}_1=\left(u_x\right)\widehat{i}+\left(u_v-g\right)\widehat{j}$,

 and two seconds later from the time of projection, it will  be ${\overrightarrow{v}}_2=\left(u_x\right)\widehat{i}+\left(u_y-2g\right)\widehat{j}$. 

According to the given condition, we have $\frac{u_y-g}{u_y}=\tan {45}^{\circ }\text{ and }{u}_y-2g=0$

Solving the above equations, we get $u_x=10\mathrm{m}/\mathrm{s}\text{ and }{u}_y=20\mathrm{m}/\mathrm{s}$

Hence, the projection speed is $\sqrt{u_x^2+u_y^2}=10\sqrt{5}\mathrm{m}/\mathrm{s}$.