We know however, that the body moves with constant acceleration of w along the -ive y-axis, this means that,
Now we substitute (2) in (1b) to obtain,
What (3) means is that, since there is no acceleration along the x-axis, the component of velocity of the point along the x-axis will remain constant at the value given by (3).
Now we can determine the value of dy/dt from (1a) as,
The velocity of the particle at the origin is thus given by setting x=0, y=0 as,
2 comments:
i have got another solution without using calculus
please tell me whether my aproach is correct or not
y=ax-bx^2
let the horizontal component of its velocity at the origin be Vx and vertical component of velocity at the origin be Vy
as the acceleration is along the negetive y direction
motion along the x direction is uniform. acceleration=-ω
eqn on motion along x direction
x=Vxt
along y direction
y=Vyt-1/2ωt^2
we know that
y=ax-bx^2
substituting for x we get y=a(Vxt)-b(Vxt)^2
equating we get
Vx=aVy
bVx^2=1/2ω
using these eqns to fing root of(Vx^2 + Vy^2)
which is the velocity at origin
= root of [(1+a^2)ω/2b ]
This is also a valid way of doing it :-)
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