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

Question

Vikas TU · Answer

Maximum Height of a projectile is given as: H = u^2sin^2(thetha)/g but to be maximum thetha should be 90 degree. And therefore, H = 64*64/2g = > 204.8 meter.

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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

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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

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