Projection at an angle - throwing stones

[Oomph!]

Hello everyone. I have this problem:

1. The problem statement, all variables and given/known d
I throw stone from height h at an angle α, then at an angle β (so I know h, α, β).
I throw it with same velocity v. The stones fall in same distance d.
The question is: find the d (distance) and v (initial velocity).

Homework Equations

So, this is projection at an angle. Coordinates of velocity and coordinates of position:

v_x=v₀cosα
v_y=v₀sinα-gt

x=tv₀cosα
y=h+tv₀sinα-0,5gt²

The Attempt at a Solution

We can make the formula for the calculation of d:

d=(v²sin2α+sqrt(v⁴*(sin2α)²+8hgv²*(cosα)²)/2g

I could solve this equation for two angles (distance is same) and find v:

(v²sin2α+sqrt(v⁴*(sin2α)²+8hgv²*(cosα)²) =
(v²sin2β+sqrt(v⁴*(sin2β)²+8hgv²*(cosβ)²)

I have done it. However, the solution is not nice, really big formula. So, I think this is not good process. There must be something easier. I know the answer:

d=h*tan(α+β)
v²=0,5gh*sin(α+β)*tan(α+β)/(cosα*cosβ)

My solution was really difficult and there was not any way how to make it easier.
So, could you show me how to make better solution?

Thank you.

 

[mfb]

The calculation is messy, but you can solve your equation for v². I would introduce a new parameter $c=\frac{2gh}{v^2}$, that simplifies the equation to
$$\sin(\alpha)\cos(\alpha) + \cos(\alpha) \sqrt{\sin^2(\alpha) - c} = \sin(\beta)\cos(\beta) + \cos(\beta) \sqrt{\sin^2(\beta) - c}$$
Note that I get a different prefactor for c.

 

[jeffbruma]

Consider the times it took to fall. You should get a quadratic equation in $t$ with two possible solutions.

 

[Oomph!]

Thank you for new parameter. Yes, it is better, but I think it is still difficult to solve it, if i want the solution in this form: v2=0,5gh*sin(α+β)*tan(α+β)/(cosα*cosβ).
However, if I use Wolfram Alpha, it works, I have the solution in this form. So, there is not a simplier way?

How can I consider the times? It is differrent in each fall.

 

[mfb]

So, there is not a simplier way?

I don't see one.

How can I consider the times? It is differrent in each fall.

I think @jeffbruma was commenting on an earlier step you did already (on the way to find d).

 

[Oomph!]

OK. Thank you very much! :)