## Engineering Mechanics: Statics & Dynamics (14th Edition)

$378m/s$
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: $v_p=\sqrt{(320.75)^2+(200)^2}=378m/s$