Answer
$378m/s$
Work Step by Step
We can determine the required velocity as follows:
$v_{p,y}=v_{p/c}\cdot sin(\phi)$
$\implies v_{p,y}=400(0.5)=200m/s$
We calculate the velocity of the projectile in the x-direction
$0+0=m_pv_{p,x}-m_cv_c$
$\implies v_c=\frac{20}{250}v_c=0.08v_c$
Now $v_{p,x}=-v_c+v_{p/c}\cdot cos(\phi)$
$v_{p,x}=\frac{v_{p/c}\cdot cos(\phi)}{1+0.08}=320.75m/s$
The velocity of the projectile is given as
$v_p=\sqrt{v_{p,x}^2+v_{p,y}^2}$
We plug in the known values to obtain:
$v_p=\sqrt{(320.75)^2+(200)^2}=378m/s$
