Question

A bomb is dropped by a fighter plane flying horizontally. To an observer sitting in the plane, the trajectory of the bomb is a :

• A. parabola in the direction of motion of plane
• B. straight line vertically down the plane
• C. parabola in a direction opposite to the motion of plane
• D. hyperbola

Hint:

In problems of relative motion, always identify the frame of reference. For an object dropped from a vehicle moving with constant velocity (neglecting air resistance), the object's horizontal velocity remains the same as the vehicle's. Thus, from the vehicle's perspective, there is no horizontal motion, only vertical motion.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to determine the path of a bomb, as seen by an observer in the plane from which it was dropped. The plane is flying horizontally. We will neglect air resistance.
Step 2: Key Formula or Approach:
This is a problem of relative motion. We need to analyze the motion of the bomb in the frame of reference of the airplane.
Let \( \vec{v}_{b,g} \) be the velocity of the bomb relative to the ground.
Let \( \vec{v}_{p,g} \) be the velocity of the plane relative to the ground.
The velocity of the bomb relative to the plane is \( \vec{v}_{b,p} = \vec{v}_{b,g} - \vec{v}_{p,g} \).
Step 3: Detailed Explanation:
Let the plane be flying with a constant horizontal velocity \( v_x \hat{i} \). So, \( \vec{v}_{p,g} = v_x \hat{i} \).
At the moment the bomb is dropped (t=0), it has the same horizontal velocity as the plane. Its initial velocity relative to the ground is \( \vec{v}_{b,g}(0) = v_x \hat{i} \).
After being dropped, the bomb is subject to gravity. Its velocity at any time t is:
- Horizontal component: \( v_{bx}(t) = v_x \) (constant, as there is no horizontal acceleration).
- Vertical component: \( v_{by}(t) = -gt \) (due to gravity, taking downward as negative).
So, the velocity of the bomb relative to the ground is \( \vec{v}_{b,g}(t) = v_x \hat{i} - gt \hat{j} \).
The velocity of the plane relative to the ground remains constant: \( \vec{v}_{p,g}(t) = v_x \hat{i} \).
Now, we find the velocity of the bomb relative to the plane:
\[ \vec{v}_{b,p}(t) = \vec{v}_{b,g}(t) - \vec{v}_{p,g}(t) \] \[ \vec{v}_{b,p}(t) = (v_x \hat{i} - gt \hat{j}) - (v_x \hat{i}) = -gt \hat{j} \] The result \( \vec{v}_{b,p}(t) = -gt \hat{j} \) shows that, in the plane's frame of reference, the bomb has no horizontal velocity component. Its only motion is a downward acceleration.
Therefore, to the observer in the plane, the bomb appears to fall straight down vertically.
Step 4: Final Answer:
The trajectory of the bomb as seen by an observer in the plane is a straight line vertically down. This corresponds to option (B). (Note: For an observer on the ground, the trajectory would be a parabola).