Animated Solution for Physics - Kinematics: A bomb is dropped by fighter plane flying horizontally. To an observer sitting in the plane, the trajectory of the bomb is a
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Visualized Solution
Initial Setup
Let the velocity of the fighter plane be vp
vp=vxi^
Velocity of Plane
The plane moves horizontally with constant velocity.
Dropping the Bomb
Velocity of the bomb with respect to ground: vb
vb=vxi^−vyj^
Components of Bomb’s Velocity
Horizontal component: vxi^ (due to inertia)
Vertical component: −vyj^ (due to gravity)
Relative Velocity Setup
Relative velocity of bomb w.r.t plane: vbp
vbp=vb−vp
Calculating Relative Velocity
vbp=(vxi^−vyj^)−vxi^
vbp=−vyj^
Conclusion for Plane Observer
Since vbp has only a vertical component,
the trajectory is a straight line vertically downwards.
The Ground Perspective
For a ground observer, the trajectory is a parabola.
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The Sigma Insight: Relative Velocity
Solution Diagram
The Relativity of Falling Bombs
Imagine you are a pilot flying a fighter plane horizontally at a constant, blistering speed. You press a button, and a bomb is released from the bay doors. What happens next? How does the bomb fall? The answer depends entirely on who is asking the question.
This classic physics problem beautifully illustrates the concept of relative velocity and the power of frames of reference.
The Ground Observer's View
Let's first look at this from the perspective of someone standing on the ground. When the bomb is released, it doesn't just drop straight down. Because of inertia, it retains the horizontal velocity of the plane at the exact moment of release.
Let's define the velocity of the plane as vp=vxi^.
The moment the bomb is dropped, its initial velocity with respect to the ground is also vxi^. However, gravity immediately begins to pull it downwards, giving it an increasing vertical velocity component, −vyj^.
So, the velocity of the bomb with respect to the ground at any time t is:
vb=vxi^−vyj^
Because the bomb is moving forward at a constant speed while accelerating downwards, it traces out a curved path. To the ground observer, the trajectory is a perfect parabola.
The Pilot's View
Now, let's jump back into the cockpit. The pilot is moving horizontally at the exact same speed as the bomb's horizontal component. To find out what the pilot sees, we must calculate the relative velocity of the bomb with respect to the plane.
The formula for relative velocity is:
vbp=vb−vp
Let's substitute our vectors into this equation:
vbp=(vxi^−vyj^)−vxi^
Notice the magic of algebra here! The horizontal components perfectly cancel each other out:
vbp=−vyj^
The Final Conclusion
The relative velocity vector vbp has absolutely no horizontal component (i^). It only has a vertical component pointing downwards (−j^).
What does this mean physically? It means that from the pilot's perspective, the bomb is not moving forward or backward at all. It is simply falling straight down, directly beneath the plane. Therefore, to an observer sitting in the plane, the trajectory of the bomb is a straight line vertically down the plane.
This problem is a fantastic reminder that motion is always relative. A parabola to one person can be a straight line to another, and both are completely correct in their own frames of reference!