Two objects are thrown at the same moment from the same…

Two objects are thrown at the same moment from the same point on the ground with the same speed $v_0 = 30\ \mathrm{m/s}$: one vertically upward, the other at $\theta = 45^\circ$ to the horizontal. What is the distance between them $t = 2\ \mathrm{s}$ later?

Explanation of Method

Distance between two objects thrown at different angles.

Step 1: Positions after 2 s

Object 1 (straight up): $x_1 = 0$.

\[ y_1 = v_0 t - \tfrac12 g t^2 = 60 - 19.62 = 40.38\ \mathrm{m}. \]

Object 2 (at 45°), with $v_0\cos 45^\circ = v_0 \sin 45^\circ \approx 21.21\ \mathrm{m/s}$: $x_2 = 21.21 \cdot 2 = 42.43\ \mathrm{m}$.

\[ y_2 = 42.43 - 19.62 = 22.81\ \mathrm{m}. \]

Step 2: Distance

\[ d = \sqrt{(42.43 - 0)^2 + (22.81 - 40.38)^2} = \sqrt{1800.3 + 308.7}. \]

\[d \approx 45.9\ \mathrm{m}\]

Shortcut: gravity cancels

Both objects fall by the same $\tfrac12 g t^2$, so gravity does not change the distance between them. It is as if they moved in straight lines from the same point with speed $v_0$ at an angle of $45^\circ$ to each other:

\[ d = 2 v_0 t \sin\frac{45^\circ}{2} = 2 \cdot 30 \cdot 2 \cdot 0.3827 \approx 45.9\ \mathrm{m}\ \checkmark \]

$y\ (\mathrm{m})$ $x\ (\mathrm{m})$ $d \approx 45.9\ \mathrm{m}$ $45^\circ$ $60\ \mathrm{m}$ $\text{without gravity}$ $-19.6\ \mathrm{m}$ $(0,\ 40.4)$ $(42.4,\ 22.8)$ $d = 2 v_0 t \sin 22.5^\circ$

thrown straight up
thrown at 45°
without gravity

Positions after 2 s (drawn to scale). Without gravity both objects would be 60 m from the start along their directions (dashed). Gravity pulls both down by the same 19.6 m, so the red segment between them is just that dashed segment shifted down.

Result

\[\boxed{\,d \approx 45.9\ \mathrm{m}\,}\]

Remark

Note. The original result, 45.88 m, differs slightly because $42.43^2$ was taken as 1796.33 instead of 1800.3. The distance grows in proportion to time: after 4 s it would be about 91.8 m (as long as both are still in the air). Compare with Translational Kinematics 025.