If a projectile is shot at an angle of #(2pi)/3# and at a velocity of #64 m/s#, when will it reach its maximum height?

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If a projectile is shot at an angle of #(2pi)/3# and at a velocity of #64 m/s#, when will it reach its maximum height?

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Answer 1 · 2016-03-11T03:48:27

#~~5.54s#

Explanation:

velocity of projection,#u=64ms^-1#
angle of projection,#alpha=2pi/3#
if time of reaching maximum height be t
then it will have zero velocity at the peak.
So#0=u*sinalpha- g*t#
#=>t=u*sinalpha/ g=64*sin(2pi/3)/10=6.4*sqrt3/2=3.2*sqrt3m~~5.54s#

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If a projectile is shot at an angle of #(2pi)/3# and at a velocity of #64 m/s#, when will it reach its maximum height?

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