Answer
$v_B= 21.905m/s$, $\theta=21^{\circ}$
Work Step by Step
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:
$\theta=21^{\circ}$
