Solution

Engineering Mechanics (Dynamics)

His speed is:

\begin{align*} \vec{v}_m &= \vec{v}_r + \vec{v}_{m/r} \\ v_{m,x} \vec{i} + v_{m,y} \vec{j} &= 2\vec{i} + 4 \sin\theta \vec{i} + 4 \cos\theta \vec{j} \\\\ &\text{Note: Angle $\theta$ is calculated from the vertical axis.}\\ v_m \cdot \frac{30}{50} \vec{i} + v_m \cdot \frac{40}{50} \vec{j} &= 2\vec{i} + 4 \sin\theta \vec{i} + 4 \cos\theta \vec{j} \\ 0.6 v_m &= 2 + 4 \sin\theta \\ 0.8 v_m &= 4 \cos\theta \quad \Rightarrow \quad v_m = \frac{4 \cos\theta}{0.8} = 5 \cos\theta. \\\\ 0.6 \cdot 5\cos\theta &= 2 + 4 \sin\theta \\ 3 \cos\theta &= 2 + 4 \sin\theta \bigg/^2 \\ 9 (1-sin^2\theta) &= 4 + 16\sin\theta + 16\sin^2\theta\\ 25 \sin^2\theta + 16\sin\theta -5 &= 0 \bigg/ : 25\\ \sin^2\theta + 0.64\sin\theta - 0.2 &= 0\\ \sin\theta_{1,2} &= \frac{-0.64 \pm \sqrt{0.4096 + 0.8}}{2} \\ &= \frac{-0.64 \pm 1.09982}{2} \\ \sin\theta_1 = -0.8699 \quad & \theta_1 = -60.45\text{\textdegree} \quad \, \, \text{We discard this solution.}\\ \sin\theta_2 = 0.2299 \quad & \theta_2 = 13.29\text{\textdegree}.\\\\ v_m &= 5\cos\theta \\ &= 5 \cdot 0.97\\ v_m &= 4.87 \, \text{ft/s.} \end{align*}

Answer

v_{m} = 4.87

ft/s.

t=10.27