Help with oblique motion question (initial height)

[jecg]

Homework Statement

How to calculate initial height (h) in an oblique motion?
Data: initial velocity (v₀) = 10 m/s
angle (α) = 45º
g = 10 m/s²
Variables: x - position in the x-axis
y - position in the y axis
r - range
h_max - maximum height
t_f - time of flight

Homework Equations

vx = vy = 5√2 m/s
y = h + tan(α) x - (g/v_0x²) x² → trajectory equation
v_y² = v_0y² - 2gΔy

The Attempt at a Solution

1^st - 0 = h + v_0y t_f - ½ g t_f² ⇔ t_f = (v_0y + √(v_0y² + 2hg))/g
in the problem t_f = (5√2 + √(50+20h))/10
and using that:
0 = h + 5√2 × ((5√2 + √(50+20h))/10) - 5×((5√2 + √(50+20h))/10)² ⇔ 0 = 0 this is redundant becuase I'm using the same formula twice, I understand the result
2^nd - using the trajectory equation:
0 = h + x - (10/ (2v_0x²) x² ⇔ x = 5 + 5√(1 - (2/5)h)
in this case x = r so, r = v_0x t_f ⇒ t_f = (1 + √(1 - (2/5)h)/√2
so I picked the expression from the 1^st attempt and matched with this new one, like this:
(5√2 + √(50+20h))/10 = (1 + √(1 - (2/5)h)/√2 ⇔ h = 0 even though this atempt gave me a "real" result, h ≠ 0 so it can't be
3^rd - in a crazy last atempt I used the expressin for t_f from second atempt and substitute it in the equation for y position:
0 = h + 5√2 × ((1 + √(1 - (2/5)h)/√2 ) - 5((1 + √(1 - (2/5)h)/√2)² ⇔ h = 0 as expected the result was 0
4^th - I also tried working with torricelli's equation, but that led me no where, I obtained this, which might be helpful: h_max = h + 5/2

 

[Simon Bridge]

"oblique motion" can refer to a lot of things ... I'm guessing this is a ballistics problem?
i.e. a projectile is fired at speed 10m/s at an angle 45deg to the horizontal - neglect air resistance.
Is that correct?

If so then

vx = vy = 5√2 m/s

... this is incorrect.