A particle moves in the plane XY with constant acceleration a directed along he negtive Y-axis. The equation of motion of the particle has the form y=Ax-Bx^(2), where A and B are positive constants. Find the velocity of the particle at the origin of the coordinates.

A particle moves in the plane XY with constant acceleration a directed

1.	A particle moves in the plane XY with constant acceleration a directed along he negtive Y-axis. The equation of motion of the particle has the form y=Ax-Bx^(2), where A and B are positive constants. Find the velocity of the particle at the origin of the coordinates.
Answer» Solution :Taking X-axis HORIZONTAL and Y-axis vertical. We GAVE `Y=Ax-Bx^(2)` COMPARING with equation of trajectory. `y=x tan alpha-(GX^(2))/(2u^(2)cos^(2)alpha)` `=x tan alpha-(gx^(2))/(2u^(2))(1+tan^(2)alpha)` We have `=tan alpha` and `B=g/(2u^(2))(1+A^(2))` (or) `u^(2)=((1+A)^(2)g)/(2B)` or `u=sqrt(((1+A)^(2)g)/(2B))`