256bits said:
Initially, I would just assume that the foam moves straight back and hits the wing and treat the problem as 1-D as a first approximation.
Since the shuttle is shaped somewhat as a triangle in one plane, the air has to move sideways as well as towards the rear along the surface of the shuttle, so it has an x and a y component and a resultant with somewhat greater velocity. A second approximation would determine the resultant air speed assuming a traingle given by the dimensions of the shuttle and assume the foam will be induced to follow the same path. Except that we are not given the dimensions of the shuttle, so from the picture one could use the 78 ft and from the top view, estimate the detach point and contact point by the transfer of the arrow, if the arrow head is the contact point. Length of arrow is 56 ft, and one should be able to work out the other dimensions of the foam motion, taking into account also that it appears to follow a path towards the top of the shuttle. That is the only reason I can see that they gave you the wingspan.
Compare that to your first estimate of the foam velocity wrt the shuttle at contact.( What is that ej thing mean, especially e )
That should make sense, yes. From an observer, on the ground, at detachment the foam is moving upwars at the same speed as the shuttle. Gravity and drag should both be acting downwards, slowing the foam down in its ascent. From an observer on the shuttle, gravity and drag are both accelerating the foam downwards. Make sense.
One way to "solve" the problem is to assume a constant drag force ( from the 2500 ft/sec velocity ) just to see where you are at with the minimum time to impact and maximum velocity at impact.
And then go on to the differential equation , which DEvans suggested. He should be able to help you with setting up.
Okay, let's simplify the problem to 1-dimension and assume that drag due to air flow is constant. Then in the first instant after the foam comes loose from the shuttle the force ## F_0 ## acting on the foam can be modeled as
$$ F_0 = mg + \frac{1}{2} \rho (2500 \textrm{ ft/s})^2 C_D A. $$
But as the foam starts to speed towards the Earth it's velocity relative to the shuttle ## v ## starts to be non-negligible and as a consequence drag starts to act also in the direction away from the earth, so for the time under consideration we could model the forces acting on the foam as
$$ F = mg + \frac{1}{2} \rho (2500 \textrm{ ft/s})^2 C_D A - \frac{1}{2} \rho v^2 C_D A. $$
Terminal velocity ## v_f ## is characterised by the following equation,
$$ 0 = g + \frac{1}{2} \rho (2500 \textrm{ ft/s})^2 \frac{C_D A}{m} - \frac{1}{2} \rho v_f^2 \frac{C_D A}{m}. $$
Which can be solved to get the terminal velocity as follows,
$$ v_f = \sqrt{\frac{2(mg + F_a)}{C_D A \rho}} = \sqrt{\frac{2 \cdot 21554\textrm{ lb ft/s}^2}{C_D A \rho}} = 1770\textrm{ ft/s}, $$
where ## F_a ## is the force due to air flow. This allows us to express the acceleration as a function of velocity introducing the terminal velocity as
$$ a(v) = 10777 \textrm{ ft/s}^2 \left[1 - \left( \frac{v}{v_f} \right)^2 \right], $$
where ## 10777 \textrm{ ft/s}^2 ## is ## (mg + F_a)/m ##.
Now, using the relationship
$$ t(v) = t_0 + \int_{v_0}^v \frac{dv}{a(v)} $$
we can write
$$ t(v) = \frac{1}{10777 \textrm{ ft/s}^2} \cdot \int_{0}^v \frac{dv}{1 - (v/v_f)^2} = \frac{v_f}{2 \cdot 10777 \textrm{ ft/s}^2} \cdot \ln \left( \frac{v_f + v}{v_f - v}\right), $$
which can be solved for ## v ## to obtain
$$ v(t) = v_f \cdot \frac{e^{2 \cdot \left( 10777 \textrm{ ft/s}^2 \right) \cdot t / v_f} - 1}{e^{-2 \cdot \left( 10777 \textrm{ ft/s}^2 \right) \cdot t / v_f} + 1}. $$
We could even obtain the velocity as a function of the traveled distance from the relationship
$$ s(v) = s_0 + \int_{v_0}^v \frac{v \cdot dv}{a(v)} $$
by writing
$$ s(v) = \frac{1}{10777 \textrm{ ft/s}^2} \int_0^v \frac{v \cdot dv}{1 - (v/v_f)^2} = - \frac{v_f^2}{2 \cdot 10777 \textrm{ ft/s}^2} \cdot \ln \left( 1 - (v/v_f)^2 \right). $$
Which can be solved for ## v ## to obtain
$$ v(s) = v_f \sqrt{1 - e^{2 \cdot \left( 10777 \textrm{ ft/s}^2 \right) \cdot s / v_f^2}}. $$
Finally we can evaluate this last function for the flight path length of the foam in the air stream,
$$ v(56 \textrm{ ft}) = 1001 \textrm{ ft/s}. $$
But the answer given to the velocity of the foam when it strikes the wing is about 840 fps, which means that something is wrong with this model.
