Projectile motion, harder

[dragonsdemesne]

Spoiler for Hidden:

If there isn't any air resistance, then the maximum height will be reached at the halfway point between O and Q.  (this isn't too hard to prove, and it should also make sense)  This means that you can find the height from a right angle triangle. (which is half of the equilateral triangle OPQ Leodip mentioned) All you need for that is Pythagoras' Theorem, which is D^2 + E^2 = F^2 (using D, E, F so as not to confuse with other labels)  F is the length of the line OP (or OQ) and I call that X.  One of D and E is the unknown height P and the other is half of X.  So you have

P^2 + (0.5X^2) = X^2

and you end up with

height = P = X sqrt(3) / 2

Using the fact that when the projectile is at point P, it has zero vertical velocity, you can take the basic equation

distance = initial velocity * time + 0.5 * acceleration * time^2

The initial velocity, as I just said, is zero, so you can ignore that term, and solving for time, you will get that

time = sqrt (2 * distance / acceleration)  (note that this is the time to fall from P to Q)

The distance is the height I just figured out, sqrt(3) / 2, and the acceleration is the acceleration of gravity, since this is only considering the vertical height, velocity,
and acceleration.  So if you substitute in the height from above, the time that the projectile takes to fall from P to Q is:

t = sqrt ( sqrt(3) X/g )

This is only half of the time that the projectile is in the air, though, because it spent an equal amount of time going from O to P.  So the entire journey's duration, T, is double this time, which is

T = 2t = 2 sqrt (sqrt(3) X/g)

The horizontal velocity Vh can be calculated from distance = velocity * time, or velocity = distance / time

Vh = distance / time

The distance is defined by the problem as X.  The time is T.

Vh = X / (2 sqrt (sqrt(3) X/g) )

The vertical distance Vv can be calculated from velocity = distance / time as well. 
This time, the distance is the height P calculated earlier.

Vv = distance / time = ( X sqrt(3) / 2 ) / ( 2 sqrt (sqrt(3) X/g) )

The velocity, v, that Leodip wants to solve for is found by taking both components into consideration.  The equation is:

v^2 = (Vh)^2 + (Vv)^2

v^2 = ( X / (2 sqrt (sqrt(3) X/g) ) )^2 + (( X sqrt(3) / 2 ) / ( 2 sqrt (sqrt(3) X/g) ))^2

Because this is butt ugly, doing some algebraic simplification gives

v^2 = 7xg / (16 sqrt(3) )

v = sqrt ( 7xg / (16 sqrt(3) ) )

That's one of the things Leodip wanted to find.

I'll have to think on the angle later; I don't have time to finish it right now.  Feel free to ask questions or point out mistakes, but it'll be a few hours at least before I check in again.

[CuCN]

The vertical distance Vv can be calculated from velocity = distance / time as well. 

The vertical velocity isn't constant, so this doesn't apply.

v = sqrt ( 7xg / (16 sqrt(3) )

Since x wasn't fixed in the problem, this doesn't set a constraint on v. This just shows that x is dependent on v, but v can still take any positive value.

[dragonsdemesne]

Actually, that's a valid point.  I think that would only be the initial velocity, then. (or the final velocity as it lands; both will be equal)  X was fixed in the problem by Leodip as OP=OQ, though.  The vertical velocity is going to vary with time due to gravity.  I also assumed that there was no horizontal acceleration, but the problem does not specify this either way.  If there -is- a nonzero horizontal acceleration, I'm not sure that this can be done.

As for the angle A, if my  initial horizontal and vertical velocities' values are correct, then it can be calculated just by using those and trigonometric ratios.  It would be tan a = Vv / Vh.  This will be true even if my Vv and Vh are totally wrong; it just means those values won't be the ones I claimed they were.  I'm pretty sure my Vh is correct, but because of CuCN's points, I think I probably made a mistake in Vv, but I'm not going to try and figure it out tonight :p  Tomorrow, after my poor little brain has rested, it should be possible to calculate the final velocity Vv as it impacts Q from height, time, and gravity.  In fact, I think it would just be v = sqrt (2gh) where g=acceleration of gravity and h=height.  Then, the height was already calculated as X sqrt(3) / 2.  I'll scribble it out tomorrow and see if that gives the same thing as the Vv I worked out today.  If it does, I think that Vv would be valid at the points P and Q, and that's all you need to figure out the angle A. 

[Sqantic Pilau]

Although this is an old topic and you've probably lost interest (or matriculated) I feel compeled to comment, as everyone seems to be missing a very simple solution.  The points O , P, and the point vertically below P form a right angled triangle.
A 1, 2, root(3) triangle in fact - so simply by inspection one sees a=60 degrees (assuming it's defined relative to the horizontal) !!!
As often is the case with school level maths - draw a diagram and think about it briefly -  the solution will often present itself.

[CuCN]

Although this is an old topic and you've probably lost interest (or matriculated) I feel compeled to comment, as everyone seems to be missing a very simple solution.  The points O , P, and the point vertically below P form a right angled triangle.
A 1, 2, root(3) triangle in fact - so simply by inspection one sees a=60 degrees (assuming it's defined relative to the horizontal) !!!
As often is the case with school level maths - draw a diagram and think about it briefly -  the solution will often present itself.

The fact that OPQ is an equilateral triangle was not in doubt, but the problem with concluding that a must be 60 degrees is that the projectile does not travel in a straight line from O to P. It is affected by gravity, so it has to be launched at an angle higher than 60 degrees in order to reach P. The question was how to find that angle.

[Leodip]

Although this is an old topic and you've probably lost interest (or matriculated) I feel compeled to comment, as everyone seems to be missing a very simple solution.  The points O , P, and the point vertically below P form a right angled triangle.
A 1, 2, root(3) triangle in fact - so simply by inspection one sees a=60 degrees (assuming it's defined relative to the horizontal) !!!
As often is the case with school level maths - draw a diagram and think about it briefly -  the solution will often present itself.

I've actually solved this, and also considered the equilateral triangle stuff to check my calculations (a>60°), but still, thanks for having tried.