- #1
jaumzaum
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An object is thrown of a given H height by a given Vo initial velocity. Calculate the biggest horizontal distance the object can reach from the starting point.
It was a questioon from the national physics olympiad in my country.
I tried to solve, but I was not able toif the object is thrown by a [itex]/alpha[/itex] angle, we can easy calculate
Horizontal Velocity [itex]Vx = Vo cos(\alpha)[/itex]
Vertical Velocity [itex]Vy = Vo sin(\alpha)[/itex]
The height in function of time
[itex]h=H+Vo sin(\alpha) t - (1/2)gt²[/itex]
If h = 0, Solving we get
[itex]t = \frac{Vo sin(\alpha) + \sqrt{ Vo²sin²(\alpha) + 2gH} )}{g} [/itex]
So horizontal distance A =
[itex]A = Vo cos(\alpha) (\frac{Vo sin(\alpha) + \sqrt{ Vo²sin²(\alpha) + 2gH} )}{g} )[/itex]Derivating this and equaling to 0 we get the \alpha for max velocity, but I've tried this and I got a 6th power equation, even Wolfram could not solve it
Do you think there's a easier way or the problem should be canceled ?
It was a questioon from the national physics olympiad in my country.
I tried to solve, but I was not able toif the object is thrown by a [itex]/alpha[/itex] angle, we can easy calculate
Horizontal Velocity [itex]Vx = Vo cos(\alpha)[/itex]
Vertical Velocity [itex]Vy = Vo sin(\alpha)[/itex]
The height in function of time
[itex]h=H+Vo sin(\alpha) t - (1/2)gt²[/itex]
If h = 0, Solving we get
[itex]t = \frac{Vo sin(\alpha) + \sqrt{ Vo²sin²(\alpha) + 2gH} )}{g} [/itex]
So horizontal distance A =
[itex]A = Vo cos(\alpha) (\frac{Vo sin(\alpha) + \sqrt{ Vo²sin²(\alpha) + 2gH} )}{g} )[/itex]Derivating this and equaling to 0 we get the \alpha for max velocity, but I've tried this and I got a 6th power equation, even Wolfram could not solve it
Do you think there's a easier way or the problem should be canceled ?