A projectile is fired at an angle of  ${30}^{\mathrm{o}}$ to the horizontal such that the vertical component of initial velocity is $80\mathrm{m}/\mathrm{s}.$  its time of fight is T. Its velocity at  $\mathrm{t}=\mathrm{T}/4$ has a magnitude of nearly

$\begin{array}{l}given angle of projection =\theta ={30}^0; vertical component of initial velocity =u_y=80 m/s;\\ we know, \frac{u_x}{u_y}=\cot {30}^{\circ }\\ \frac{u_x}{80}=\sqrt{3}\\ u_x=80\sqrt{3}m/ s\\ time of flight= T=\frac{2u_y}{g}=\frac{2\times 80}{10}\\ \therefore T=16s\\ \text{ At }t=\frac{T}{4}=4s,\\ u_x=v_x=80\sqrt{3}m/s\\ v_y=u_y-gt substitute obtained values,\\ v_y=80-10\times 4=40\mathrm{m}/\mathrm{s}\\ \left|v\right|=\sqrt{{v_x}^2+{v_y}^2}\\ \therefore v=\sqrt{(80\sqrt{3}{)}^2+(40{)}^2}\approx 144\mathrm{m}/\mathrm{s}\end{array}$