An object is thrown from a seaside cliff with speed v_0 = 10…

An object is thrown from a seaside cliff with speed $v_0 = 10\ \mathrm{m/s}$ at $\alpha = -30^\circ$ (30° below the horizontal). Find its position and velocity $t = 2\ \mathrm{s}$ later.

Explanation of Method

An object thrown 30° downward from a cliff.

Step 1: Position

\[ x = v_0\cos\alpha\,t = 10 \cdot 0.866 \cdot 2 \approx 17.3\ \mathrm{m}, \]

\[ y = v_0\sin\alpha\,t - \tfrac12 g t^2 = -10 - 19.62 = -29.6\ \mathrm{m}. \]

\[(x, y) \approx (17.3\ \mathrm{m},\ -29.6\ \mathrm{m})\]

Step 2: Velocity

\[ v_x = 8.66\ \mathrm{m/s}, \]

\[ v_y = -5 - 19.62 = -24.62\ \mathrm{m/s}, \]

\[ v = \sqrt{8.66^2 + 24.62^2}. \]

\[\begin{gathered} v \approx 26.1\ \mathrm{m/s} \\ \text{at } 70.6^\circ \text{ below the horizontal} \end{gathered}\]

Check

$ v^2 = v_0^2 + 2g\,|y| = 100 + 581.2 $

$ v^2 = 681.2 \Rightarrow v \approx 26.1\ \mathrm{m/s}\ \checkmark $

$v_0 = 10\ \mathrm{m/s},\ -30^\circ$ $(17.3,\ -29.6)\ \mathrm{m}$ $v \approx 26.1\ \mathrm{m/s},\ 70.6^\circ \downarrow$ $17.3\ \mathrm{m}$ $29.6\ \mathrm{m}$

Drawn to scale (velocity arrows 2× in m/s). Thrown 30° downward, the object after 2 s is 17.3 m out and 29.6 m down, moving steeply downward.

Result

\[\boxed{\,(x, y) \approx (17.3\ \mathrm{m},\ -29.6\ \mathrm{m})\,}\]

\[\boxed{\,\begin{gathered} v \approx 26.1\ \mathrm{m/s} \\ \text{at } 70.6^\circ \text{ below the horizontal} \end{gathered}\,}\]