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 (a) the position of the bomb $t = 10\ \mathrm{s}$ later, relative to the release point; (b) its speed at that moment; (c) the direction of its motion relative to the horizontal at that moment.
Explanation of Method
Position and velocity of a bomb 10 s after release.
Step 1: Horizontal and vertical motion
The bomb keeps the plane's horizontal speed $v_0 = 250/3.6 \approx 69.44\ \mathrm{m/s}$ and falls freely from rest vertically.
Step 2: (a) Position after 10 s
\[ x = v_0 t = 69.44 \cdot 10 = 694.4\ \mathrm{m}, \]
\[ y = -\tfrac12 g t^2 = -490.5\ \mathrm{m}. \]
\[694\ \mathrm{m} \text{ ahead of and } 490.5\ \mathrm{m} \text{ below the release point}\]
Step 3: (b) Speed
\[ v_x = 69.44\ \mathrm{m/s}, \]
\[ v_y = g t = 98.1\ \mathrm{m/s}\ \text{(down)}, \]
\[ v = \sqrt{69.44^2 + 98.1^2}. \]
\[v \approx 120.2\ \mathrm{m/s}\]
Step 4: (c) Direction
\[ \tan\theta = \frac{v_y}{v_x} = \frac{98.1}{69.44} \approx 1.413. \]
\[\theta \approx 54.7^\circ \ \text{below the horizontal}\]
- horizontal part
- vertical part (drawn half length)
- velocity
Result
\[\boxed{\,694\ \mathrm{m} \text{ ahead of and } 490.5\ \mathrm{m} \text{ below the release point}\,}\]
\[\boxed{\,v \approx 120.2\ \mathrm{m/s}\,}\]
\[\boxed{\,\theta \approx 54.7^\circ \ \text{below the horizontal}\,}\]
Remark
About the two original solutions. Both had the right method and the right speed, but they disagreed on the angle (54.33° and 55.3°); $\arctan 1.413 = 54.7^\circ$. Note also that without air resistance the bomb stays exactly below the plane the whole time.