An airplane flies horizontally at a constant speed v = 250…
An airplane flies horizontally at a constant speed $v = 250\ \mathrm{km/h}$ and releases a bomb. Find, $t = 10\ \mathrm{s}$ after the release, (a) the position of the bomb, (b) its speed, and (c) the direction of its motion relative to the horizontal. Ignore air resistance and assume the bomb has not yet reached the ground.
Explanation of Method
A bomb 10 s after release from a plane.
Step 1: Initial velocity
The bomb leaves with the plane's velocity, which is horizontal: $v_x = \frac{250}{3.6} \approx 69.44\ \mathrm{m/s}$.
\[ v_{y0} = 0. \]
Step 2: (a) Position
With the release point as origin ($x$ forward, $y$ up): $x = v_x t = 69.44 \cdot 10 \approx 694.4\ \mathrm{m}$.
\[ y = -\tfrac12 g t^2 = -\tfrac12 \cdot 9.81 \cdot 100 = -490.5\ \mathrm{m}. \]
\[(x, y) \approx (694.4\ \mathrm{m},\ -490.5\ \mathrm{m})\]
That is 694.4 m ahead of the release point and 490.5 m below it, directly under the plane.
Step 3: (b) Speed
\[ v_y = -g t = -98.1\ \mathrm{m/s}, \]
\[ v = \sqrt{69.44^2 + 98.1^2} = \sqrt{4822.5 + 9623.6}. \]
\[v \approx 120.2\ \mathrm{m/s}\]
Step 4: (c) Direction
\[ \tan\theta = \frac{|v_y|}{v_x} = \frac{98.1}{69.44} \approx 1.413. \]
\[\theta \approx 54.7^\circ \ \text{below the horizontal}\]
Check
Energy: $ v^2 = v_x^2 + 2g\,|y| = 4822.5 + 9623.6 $
$ v = \sqrt{14\,446} \approx 120.2\ \mathrm{m/s}\ \checkmark $
Result
\[\boxed{\,(x, y) \approx (694.4\ \mathrm{m},\ -490.5\ \mathrm{m})\,}\]
\[\boxed{\,v \approx 120.2\ \mathrm{m/s}\,}\]
\[\boxed{\,\theta \approx 54.7^\circ \ \text{below the horizontal}\,}\]
Remark
About the original solution. The position and the speed are right, but the angle is $\arctan 1.413 \approx 54.7^\circ$, not 55.3°. (It also has a small slip: $69.44^2 = 4822.5$, not 4827.2, which does not change the rounded speed.) The plane's altitude is not given; the answer assumes the plane is higher than 490.5 m, so the bomb is still falling after 10 s.