A body is thrown with speed v_0 = 200 m/s at α = 60^° to the…

A body is thrown with speed $v_0 = 200\ \mathrm{m/s}$ at $\alpha = 60^\circ$ to the horizontal. Find the velocity components $v_x$ and $v_y$: at the start, at the highest point, and when it hits the ground.

Explanation of Method

Velocity components at launch, top and landing.

Step 1: General rule

Horizontally there is no acceleration; vertically the acceleration is $-g$: $v_x = v_0\cos\alpha = \text{const}$.

\[ v_y = v_0\sin\alpha - gt. \]

Step 2: At the start

\[v_x = 100\ \mathrm{m/s}, \qquad v_y = 100\sqrt3 \approx 173.2\ \mathrm{m/s}\]

Step 3: At the highest point

The body stops rising, so the vertical part is zero: $v_x = 100\ \mathrm{m/s}, \qquad v_y = 0$.

Step 4: On impact (same level as the start)

The flight is symmetric, so the vertical part has the same size as at the start but points down: $v_x = 100\ \mathrm{m/s}, \qquad v_y \approx -173.2\ \mathrm{m/s}$.

Check

Flight time $ T = \tfrac{2 \cdot 173.2}{9.81} \approx 35.3\ \mathrm{s} $; $ v_y(T) = 173.2 - 9.81 \cdot 35.3 \approx -173.2\ \mathrm{m/s}\ \checkmark $

$v_x = 100$ $v_y = 173.2$ $\text{top: } v_x = 100,\ v_y = 0$ $v_x = 100,\ v_y = -173.2$

Drawn to scale (speeds in m/s). The horizontal part of the velocity is 100 m/s everywhere; the vertical part goes from +173.2 through 0 at the top to −173.2 at landing.

Result

\[\boxed{\,v_x = 100\ \mathrm{m/s}, \qquad v_y = 100\sqrt3 \approx 173.2\ \mathrm{m/s}\,}\]

\[\boxed{\,v_x = 100\ \mathrm{m/s}, \qquad v_y = 0\,}\]

\[\boxed{\,v_x = 100\ \mathrm{m/s}, \qquad v_y \approx -173.2\ \mathrm{m/s}\,}\]