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

$v_B= 21.905m/s$, $\theta=21^{\circ}$
The required speed and the angle can be determined as follows: We convert the velocity at $a$ to $m/s$: $v_A=\frac{70\times 1000}{3600}=19.44m/s$ Now, we have: $h_A=r_Atan60=60tan60=103.923m$ and $h_B=r_Btan60=57tan60=98.727m$ Now we apply the equation of conservation energy $\frac{1}{2}mv_A^2+mgh_A=\frac{1}{2}mv_B^2+mgh_B$ We plug in the known values to obtain: $\frac{1}{2}(19.44)^2+(9.81)(103.923)=\frac{1}{2}v_B^2+9.81(98.727)$ This simplifies to: $v_B=21.905m/s$ According to the conservation of angular momentum $r_Am(v_A)_t=r_Bm(v_B)_t$ We plug in the known values to obtain: $1166.4=1248.585cos\theta$ This simplifies to: $\theta=21^{\circ}$