# How to get initial and final velocity? need help thanks

CAF123
Gold Member
The way I thought of doing it initially was to use the well known trajectory equation $$y = x tanθ - \frac {g x^2}{2 v_o^2 cos^2 θ}$$ and put in values for $y$ and $x$ depending on where the human cannonball is at specific stages of the journey.
For example, we want, at say $x = 3m, y$ to be 15.50m. Here, my assumption is that 3m is the distance from the first wheel to the cannon. The only unknowns you have is $v_o$ and $θ$.

I then took another stage in the journey say, right at the end, where $y = 2.50 m$ and $x$ is about $(10.63 +6) ≈ 17m$. Thus, I have two eqns and two unknowns which can be solved to yield some $θ$ and $v_o$.

Is this method plausible? Of course, after getting some result, to get minimal velocity, you could tweak the distance from the cannon to the first wheel and see what happens.

ehild
Homework Helper
Yes, you can try and see when you get the minimum vo. But the problem has an exact solution.
The geometry of the problem is not clearly stated. One can understand that the three wheels altogether cover 10.63 m distance, or each of them does.

a human cannonball in 1940 soared over three ferris wheels, each 18 meters high covering a horizontal distance of 10.63 meters. Assuming that the point of projection is 2.50 meters above the ground and that he landed safely on a net placed at the same level, find his initial velocity.

Looking at the picture of a Ferris Wheel, I am inclined to believe the second geometry is valid.

ehild

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Is this method plausible? Of course, after getting some result, to get minimal velocity, you could tweak the distance from the cannon to the first wheel and see what happens.

Technically, it should work. In practice, it seems to result in very messy algebra. The method of looking at the energy state at the apex, as detailed by ehild, gets you to the result much quicker.

CAF123
Gold Member
I would agree, but it is nice to know there are other ways to tackle things.