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Position function, Polar coordinates

Position function, Polar coordinates

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(1)
Angle phi = p and theta = T
I will introduce an unknown integration constant after every integration and use boundary conditions to find its value.
Ball experiences only the gravitational acceleration which acts down ward.

Along the x-axis, acceleration, d^2 x/ dt^2 = - g sin (p)

Integrating once: dx/dt = - g sin (p) t + A

Find the integration constant A: Use the boundary condition that at t = 0, x component of the velocity = v0 cos (t).

vo cos (t) = 0 + A
A = vo cos T
Hence,

dx/dt = - g sin (p) t + vo cos T

Integrate one more time.

x = - g sin (p) t^2/2 + vo cos T * t + C where C is the integration constant.

Boundary condition: at t = 0 x = 0
This gives C = 0
Hence,

x = - g sin (p) t^2/2 + vo cos T * t

Do a similar analysis for y motion:

Along the y-axis, acceleration, d^2 y/ dt^2 = - g cos (p)

Integrate: dy/dt = -g cos (p) t + D
Boundary condition: t = 0 dy/dt = v0 sin T
Substituting we get, vo sin T = ...

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