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A particle is projected from the origin in X-Y plane. Acceleration of particle in Y direction is a. If equation of path of the particle is y = ax-bx^(2), then find initial velocity of the particle. |
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Answer» Solution :`y=ax- BX^(2) "" y= X tan theta= (alpha x^(2))/(2 u cos^(2) theta)` `tan theta= a and (alpha)/(2U cos^(2) theta)=b u= sqrt((alpha(1+a^(2)))/(2B)` |
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