Projectile Motion Problem. Is my answer correct ?

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SUMMARY

The projectile motion problem involves determining the launch angle required for a projectile to travel from point A to point B while just missing a pole of height H located at the tip of a double inclined plane with angles α and β. The derived formula for the height h is h = (D * tan(α) * tan(β)) / (tan(α) + tan(β)). The launch angle φ is calculated using the equation φ = tan-1(((1/2 * g * tcollide2 + H)(tan(α) + tan(β)) + D * tan(α) * tan(β)) / (D * tan(β))). The solution requires knowledge of the initial velocity (V0), incline angles (α and β), and the height of the pole (H).

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Homework Statement


A projectile is to be launched so as to go from A to B [which are respectively at the bases of a double inclined plane having angles [itex]\alpha[/itex] and [itex]\beta[/itex] as seen in the figure] and just barely miss a pole of height [itex]H[/itex] that is located at the tip. If the distance between A and B is D, find the angle with the horizontal at which the projectile should be launched.

[PLAIN]http://b1111.hizliresim.com/r/k/llf5.jpg

The Attempt at a Solution


[PLAIN]http://b1111.hizliresim.com/r/k/llds.jpg

I found [itex]h=\frac{D\tan\alpha\tan\beta}{\tan\alpha+\tan\beta}[/itex]

and I considered a particle which is at B is falling free.And Vo vector aimed to point B when t=0 , so they must collide at [itex]t=t_{collide}[/itex] and at top of the H.
Then i wrote [itex]\tan\phi=\frac{1/2gt_{collide}^2+H+h}{\frac{D\tan\beta}{\tan\alpha+\tan\beta}}[/itex]
and i found
[itex]\phi= \tan^-1(\frac{(1/2gt_{collide}^2+H)(\tan\alpha+\tan\beta)+D\tan \alpha \tan\beta}{D\tan\beta})[/itex]

Is my answer is correct ? and are there any solutions for conditions which are before reaching maximum height and minumum height ?
 
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Your "answer" assumes you know the the time it will take the projectile to reach the top of the pole.

The answer should be a function of the initial velocity (V0), the angles of the inclines (α and β), and the height of the pole from the top of the incline (H).
 
Did you find an answer?
 

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