Home Physics Motion in a Plane General If at a point of the parabolic path the velo…
Physics Motion in a Plane General Subjective Type
Published on: September 12, 2026

If at a point of the parabolic path the velocity be u and the inclination to the horizon be , at what time the particle is moving at right angles to its former direction.

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Step 1: Let's denote the initial velocity as u and the angle of inclination as \( \theta \). The components of the velocity can be expressed as:
\[ v_x = u \cos(\theta) \]
\[ v_y = u \sin(\theta) \]

Step 2: For the particle to move at right angles to its former direction, the direction of the velocity vector needs to make an angle of \( 90^{\circ} \) with its original direction.
This means that the new velocity vector's direction should be perpendicular to the initial velocity vector.

Step 3: The condition for two vectors to be perpendicular is that their dot product equals zero. Thus, if the current velocity is expressed as:
\[ \text{Current velocity} = (v_x', v_y') \]
we have:
\[ v_x' v_x + v_y' v_y = 0 \]

Step 4: If we know the equations of motion under the influence of gravity, where the velocity components change due to acceleration (gravity affects vertical velocity), we can express velocity at any time t as:
\[ v_x' = u \cos(\theta) \]
\[ v_y' = u \sin(\theta) - g t \]

By replacing these in the dot product condition, we can solve for the time t when the particle is moving perpendicular.

Conclusion: This will involve solving the resulting equations to find out the time when the condition of perpendicularity holds true. Thus, performing these calculations would lead us to the answer. Therefore, the required time can be derived from the above steps.

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