# Help with oblique motion question (initial height)

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1. Nov 16, 2014

### jecg

1. The problem statement, all variables and given/known data
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
hmax - maximum height
tf - time of flight
2. Relevant equations
vx = vy = 5√2 m/s
y = h + tan(α) x - (g/v0x2) x2 → trajectory equation
vy2 = v0y2 - 2gΔy
3. The attempt at a solution
1st - 0 = h + v0y tf - ½ g tf2 ⇔ tf = (v0y + √(v0y2 + 2hg))/g
in the problem tf = (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
2nd - using the trajectory equation:
0 = h + x - (10/ (2v0x2) x2 ⇔ x = 5 + 5√(1 - (2/5)h)
in this case x = r so, r = v0x tf ⇒ tf = (1 + √(1 - (2/5)h)/√2
so I picked the expression from the 1st 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
3rd - in a crazy last atempt I used the expressin for tf 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
4th - I also tried working with torricelli's equation, but that led me no where, I obtained this, which might be helpful: hmax = h + 5/2

2. Nov 17, 2014

### 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
... this is incorrect.