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A particle is projected with a velocity of 64m/s at angle 45 degree to the horizontal. find the highest height attained by the particle..g= 10 A particle is projected with a velocity of 64m/s at angle 45 degree to the horizontal. find the highest height attained by the particle..g= 10
Maximum Height of a projectile is given as:H = u^2sin^2(thetha)/gbut to be maximum thetha should be 90 degree.And therefore,H = 64*64/2g = > 204.8 meter.
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