A body is thrown at an angle α = 30^° to the horizontal.…

A body is thrown at an angle $\alpha = 30^\circ$ to the horizontal. After $t = 2\ \mathrm{s}$ its speed is $v = 400\ \mathrm{m/s}$. Find the initial speed $v_0$.

Explanation of Method

Initial speed from the speed after 2 s.

Step 1: Velocity components after 2 s

\[ v_x = v_0\cos 30^\circ, \]

\[ v_y = v_0\sin 30^\circ - gt = \tfrac12 v_0 - 19.62. \]

Step 2: Equation for $v_0$

\[ v^2 = v_x^2 + v_y^2 = \tfrac34 v_0^2 + \left(\tfrac12 v_0 - 19.62\right)^2 = v_0^2 - 19.62\,v_0 + 384.9. \]

(In general $v^2 = v_0^2 - 2v_0 g t\sin\alpha + g^2t^2$.) With $v = 400$: $v_0^2 - 19.62\,v_0 - 159\,615 = 0$.

\[ v_0 = \frac{19.62 + \sqrt{19.62^2 + 4 \cdot 159\,615}}{2} = \frac{19.62 + 799.28}{2}. \]

\[v_0 \approx 409.4\ \mathrm{m/s}\]

Check

$ v_x = 354.6\ \mathrm{m/s}, \quad v_y = 204.7 - 19.6 = 185.1\ \mathrm{m/s} $

$ v = \sqrt{354.6^2 + 185.1^2} \approx 400.0\ \mathrm{m/s}\ \checkmark $

$\vec v_0 \ (v_0 \approx 409\ \mathrm{m/s})$ $\vec g\,t = 19.62\ \mathrm{m/s}\ \downarrow$ $\vec v\ (v = 400\ \mathrm{m/s})$ $30^\circ$

The velocity after 2 s is the initial velocity plus $\vec g t$ (drawn 3× longer for visibility). Gravity only changes the vertical part, and after 2 s that change is small compared with the speed.

Result

\[\boxed{\,v_0 \approx 409.4\ \mathrm{m/s}\,}\]