InterviewSolution
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InterviewSolution
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Projectile motion is a combination of two one-dimensional motion: one in horizontal and other in vertical direction. Motion in 2D means in a plane. Necessary condition for 2D motion is that the velocity vector is coplanar to the acceleration vector. In case of projectile motion, the angle between velocity and acceleration will be `0^@ltthetalt180^@`. During the projectile motion, the horizontal component of velocity ramains unchanged but the vertical component of velocity is time dependent. Now answer the following questions: A particle is projected from the origin in the x-y plane. The acceleration of particle in negative y-direction is `alpha`. If equation of path of the particle is `y = ax - bx^2`, then initial velocity of the particle isA. `sqrt(alpha/(2b))`B. `sqrt((alpha(1+a^2))/(2b))`C. `sqrt(alpha/(a^2))`D. `sqrt((alphab)/(a^2))` |
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Answer» Correct Answer - B b. `y = ax - bx^2` or `(dy)/(dt) = a(dx)/(dt) - 2bx(dx)/(dt) = (a-2bx) (dx)/(dt)` Initially `x=0,v_y = av_x` `(d^2y)/(dt^2) = (a-2bx) (d^2x)/(dt^2)+[-2b(dx)/(dt)]` `= (a-2bx)(d^2y)/(dt^2) - 2b ((dx)/(dt))^2` At `t = 0 , (d^2y)/(dt^2) = -alpha, x= 0, (d^2x)/(dt^2) =0` `rArr -alpha = -2b((dx)/(dt))^2 = -2b(v_x)^2 or v_x = sqrt(a/(2b))` `rArr v = sqrt(v_(x)^(2) + v_(y)^(2)) = v_x sqrt(1+a^2) = sqrt((alpha(1+a^2))/(2b))` . |
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