Solution:
\[ tan \alpha = \frac{-gt}{u} \]
\[ tan (-60^{\circ })= \frac{-10t}{20/\sqrt{3}} \]
Solving t =2 sec
A particle is projected horizontally with a speed of 20/√3 m/s, from some height at t = 0. At what time will its velocity make 60° angle with the initial velocity
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