Suppose initial velocity is $v$ and angle of projection$\left(\theta \right)$ is given $60°.$

Given,

Vertical component of initial velocity $\left(v \sin \theta \right)=40 \mathrm{m} {\mathrm{s}}^{-1}$

$\Rightarrow v=46.19 \mathrm{m} {\mathrm{s}}^{-1}$

Time of flight $\left(T\right)=\frac{2 v \sin \theta }{g}=\frac{2\times 40}{10}=8 \mathrm{s}$

Given, time $\left(t\right)=\frac{T}{4}=2 \mathrm{s}$

Suppose vertical component of velocity after time $t$ is $v_1.$

Applying first eqaution of motion-

$v_1=v \sin 60°-gt$

$\Rightarrow v_1=40-20=20 \mathrm{m} {\mathrm{s}}^{-1}$

Horizontal component remains uncharged.

$\therefore v_H=v \cos 60°=23.09 \mathrm{m} {\mathrm{s}}^{-1}$

Magnitude of velocity after time $t \left(v_t\right)=\sqrt{v_1^2+v_H^2}$

$\Rightarrow \left|v_t\right|=\sqrt{{20}^2+23.{09}^2}=30.54 \mathrm{m} {\mathrm{s}}^{-1}$