A body of mass \( m \) is projected along a rough inclined plane (having an angle of inclination with horizontal \( \theta \), equal to angle of repose) with a velocity \( v \). It travels up a maximum distance \( s \) before it comes to a halt. Then \( v \) is:
\( \sqrt{gs \cos\theta} \)
\( 2\sqrt{gs \sin\theta} \)
\( 2\sqrt{gs \tan\theta} \)
\( \sqrt{gs \tan^2\theta} \)
Solution:
Since the inclination equals the angle of repose, \( \mu = \tan\theta \). The acceleration down the incline during upward motion is \( a = g \sin\theta + \mu g \cos\theta = 2g \sin\theta \). Using \( v^2 = 2as \), we get \( v = 2\sqrt{gs \sin\theta} \).
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