A projectile is thrown with a velocity of $10\sqrt{2}m/s$ at an angle of ${45}^{\circ }$ with horizontal. The interval between the moments when speed is  $\sqrt{125}m/s$ is  $(g=10m/s^2)$

Here,  $u=10\sqrt{2}m/s,\theta ={45}^{\circ },v=\sqrt{125}m/s$

At any instant t, velocity is given by

$$\begin{gathered} v=\sqrt{v_x^2+v_y^2}=\sqrt{{\left(u\cos \theta \right)}^2+{(u\sin \theta -gt)}^2} \\ \Rightarrow \sqrt{125}=\sqrt{{(10\sqrt{2}\times 1/\sqrt{2})}^2+{(10\sqrt{2}\times 1/\sqrt{2}-10t)}^2} \\ \Rightarrow 125=100(t^2-2t+2)\Rightarrow t^2-2t+0.75=0 \\ \therefore t_1=0.5sand{t}_2=1.5s\therefore t_2-t_2=1.0s \end{gathered}$$