Consider a particle located at (x

x = x

y = y

If t>0, the particle will reach y = 0 when

t = {v

(found using the abc formula)

If we call the time t

x

The question is: at initial height y

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 y

Thank you...

_{0}, y_{0}), having inital velocity v_{0}, 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 = x

_{0}+ v_{0}*cos[the]*ty = y

_{0}+ v_{0}*sin[the]*t - 0.5*g*t^{2}If t>0, the particle will reach y = 0 when

t = {v

_{0}*sin[the] + sqrt((v_{0}*sin[the])^{2}+2*g*y_{0})}/g(found using the abc formula)

If we call the time t

_{1}, then at that time the x position of the particle is (call it x_{1})...x

_{1}= x_{0}+ v_{0}*cos[the]*t_{1}The question is: at initial height y

_{0}, what angle will make x_{1}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 y

_{0}= 0, the problem will be easy to solve, with the answer 45 degrees.Thank you...

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