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A particle is projected with a velocity of 64m/s at angle 45 degreeto the horizontal. find the highest height attained by the particle..g= 10

Samson , 8 Years ago
Grade 9
anser 1 Answers
Vikas TU

Last Activity: 8 Years ago

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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