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  4. At the top of the trajectory of a projectile, thrown at an angle of projection θ <90° , its:

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Q. At the top of the trajectory of a projectile, thrown at an angle of projection $ \theta <90{}^\circ , $ its:

KEAMKEAM 2000

Solution:

At the highest point (P) of the project trajectory, vertical component of velocity is zero, only horizontal component exists which is perpendicular to the direction of acceleration.

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