The only trajectory problem on my whole SI sheet (trajectory) I cannot do

[realfuzzhead]

Homework Statement

This a trajectory problem. A ball is launched from ground level at a unknown angle. ( ∅_o= ? ) at an unknown velocity ( V_o = ? )

The ball travels 20 meters in the x direction and 25 meters up in the y direction. At this point (20,25) the ball hits a wall and a -45 angle below the horizon. Find the initial velocity ( Mag and direction ) and the speed when the ball hits the wall

Homework Equations

the only equations we have been given are

V_yfinal = V_o( Sin ∅_o) - gt
Δ Y = V_o( Sin ∅_o) - 1/2g (t)^2

ΔX = V_o(cos ∅) (t)

V_o( Cos ∅_o) = Vf(Cos ∅_o)

Xmax = (V_o^2 * (Sin 2∅))/ g

The Attempt at a Solution

I cannot post all my notes. I have finished, and double checked all my other answers on 2 whole trajectory work sheets. But I cannot for the life of me figure out how to find the initial angle as a function of the final angle again the change in distance.
I have even tried switching the whole problem around and shooting at 45° from 25 meters up and I still cannot find a result that works.

I have done probably 10-11 different tries of canceling out different variables (especially Vo and Vf and T) and finding them as products or sums of other variables, but I can never find a way to get the original y velocity or the orignal x velocity.

I do know that when the ball hits the wall the Y velocity is = to the -(x velocity), because tan-1 45° = -1 (-y/x)

I have spent over 2 hours on this problem and I have searched the interwbs for help but I cannot for the life of me figure out how to find the original angle if I know the final angle but don't know either the final or intial x or y velocities. I also don't know the time

I really, really don't like asking for answers to problems because I really like sitting down with a cup of joe and figuring it out for myself and actually learning, but I literally have grown some grey hairs and ruined my weekend over the amount of stress I have experienced over this problem, and I would really really appreciate if someone could answer this one if not give me a HUGE hint

 

[217 MeV]

If the ball hits the wall at an angle of 45° after traveling 25m upwards, what does that tell you about what the maximum height reached would have been if the wall was not present?

 

[NascentOxygen]

Δ Y = Vo( Sin ∅o) - 1/2g (t)^2

Should be Δ Y = Vo( Sin ∅o)t[/color] - 1/2g (t)^2

 

[PeterO]

Homework Statement

This a trajectory problem. A ball is launched from ground level at a unknown angle. ( ∅_o= ? ) at an unknown velocity ( V_o = ? )

The ball travels 20 meters in the x direction and 25 meters up in the y direction. At this point (20,25) the ball hits a wall and a -45 angle below the horizon. Find the initial velocity ( Mag and direction ) and the speed when the ball hits the wall

Homework Equations

the only equations we have been given are

V_yfinal = V_o( Sin ∅_o) - gt
Δ Y = V_o( Sin ∅_o) - 1/2g (t)^2

ΔX = V_o(cos ∅) (t)

V_o( Cos ∅_o) = Vf(Cos ∅_o)

Xmax = (V_o^2 * (Sin 2∅))/ g

The Attempt at a Solution

I cannot post all my notes. I have finished, and double checked all my other answers on 2 whole trajectory work sheets. But I cannot for the life of me figure out how to find the initial angle as a function of the final angle again the change in distance.
I have even tried switching the whole problem around and shooting at 45° from 25 meters up and I still cannot find a result that works.

I have done probably 10-11 different tries of canceling out different variables (especially Vo and Vf and T) and finding them as products or sums of other variables, but I can never find a way to get the original y velocity or the orignal x velocity.

I do know that when the ball hits the wall the Y velocity is = to the -(x velocity), because tan-1 45° = -1 (-y/x)

I have spent over 2 hours on this problem and I have searched the interwbs for help but I cannot for the life of me figure out how to find the original angle if I know the final angle but don't know either the final or intial x or y velocities. I also don't know the time

I really, really don't like asking for answers to problems because I really like sitting down with a cup of joe and figuring it out for myself and actually learning, but I literally have grown some grey hairs and ruined my weekend over the amount of stress I have experienced over this problem, and I would really really appreciate if someone could answer this one if not give me a HUGE hint

The parabolic path followed passes through 0,0 and 20,25, and the gradient at (20,25) is -1
The second derivative anywhere is -9.8

Surely if you put all that together the answer can be extracted
That enables you

 

[realfuzzhead]

Should be Δ Y = Vo( Sin ∅o)t[/color] - 1/2g (t)^2

I typed this very late last night, I promise you that that "t" was there in every single one of my equations. I lose track of variables when I have to keep typing out all the little subscripts and figures, sorry.

If the ball hits the wall at an angle of 45° after traveling 25m upwards, what does that tell you about what the maximum height reached would have been if the wall was not present?

it hits the wall at a -45 degree angle after it maximum y value

 

[realfuzzhead]

"The parabolic path followed passes through 0,0 and 20,25, and the gradient at (20,25) is -1
The second derivative anywhere is -9.8

Surely if you put all that together the answer can be extracted
That enables you"I just started calculus 2 weeks ago. I am decent with derivatives but to be honest I have never done a lick of integration, which is what I would need to go from having the derivative to the original right?

 

[cmb]

Define a time, t, for yourself, that being how long it takes to get to the wall. So we know at time t its horizontal and vertical velocities are both 20/t.

Now ask a different question - with initial velocities 20/t both horiz and vertical, from a height 25, what angle will its flight intersect the horizontal if it goes from there?