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agro
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Consider a particle located at (x0, y0), having inital velocity v0, and the angle of inclination of the velocity vector is [the]. The particle is subjected to a vertical acceleration of -g. It's equation of motion is...
x = x0 + v0*cos[the]*t
y = y0 + v0*sin[the]*t - 0.5*g*t2
If t>0, the particle will reach y = 0 when
t = {v0*sin[the] + sqrt((v0*sin[the])2+2*g*y0)}/g
(found using the abc formula)
If we call the time t1, then at that time the x position of the particle is (call it x1)...
x1 = x0 + v0*cos[the]*t1
The question is: at initial height y0, what angle will make x1 maximum?
I found it very hard to make the equation simple enough to be able to solve this problem... Can anyone help me?
PS: in the case of y0 = 0, the problem will be easy to solve, with the answer 45 degrees.
Thank you...
x = x0 + v0*cos[the]*t
y = y0 + v0*sin[the]*t - 0.5*g*t2
If t>0, the particle will reach y = 0 when
t = {v0*sin[the] + sqrt((v0*sin[the])2+2*g*y0)}/g
(found using the abc formula)
If we call the time t1, then at that time the x position of the particle is (call it x1)...
x1 = x0 + v0*cos[the]*t1
The question is: at initial height y0, what angle will make x1 maximum?
I found it very hard to make the equation simple enough to be able to solve this problem... Can anyone help me?
PS: in the case of y0 = 0, the problem will be easy to solve, with the answer 45 degrees.
Thank you...
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