then the instantaneous velocity at a height = (1/2)the maximum height,is u [(1-(sin^2 X)/2)]^(1/2) (i,e product of initial velocity and the square root of the difference of 1 and half the square of sine of the angle of projection). Now the question is that,as the horizontal velocity of a projectile is constant,the cosine component of this instantaneous velocity should give the horizontal velocity i,e
u cosX.But where does it ?I am unable to convert the expression of ins. velocity to u cosX.Please help
2006-10-23
16:39:49
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3 answers
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asked by
Anonymous