Chapter 12 - Kinematics of a Particle - Section 12.6 - Motion of a Projectile - Problems - Page 51: 87

$\theta=76^{\circ}$ $v_A=49.8$ ft/s $h=39.7$ ft

We set up the equations as follows: $s=v_0t$ $18=v_A\cos\theta(1.5)$ $v^2=v_0^2+2a_c(s-s_0)$ $0=(v_A\sin\theta)^2+2(-32.2)(h-3.5)$ $v=v_0+a_ct$ $0=v_A\sin\theta-32.2(1.5)$ $48.3=v_A\sin\theta$ $18=v_A\cos\theta(1.5)$ $12=v_A\cos\theta$ $48.3=v_A\sin\theta$ We solve for the angle: $\theta=76^{\circ}$ Now, solve for velocity and height: $v_A=49.8$ ft/s $h=39.7$ ft