Hey guys I am doing Q17 of the old fitz projectile question and it is insane.. the working out is also very rigorous so here I go:
Q: A stone is thrown so that it will hit a bird at the top of a pole. However, at the instant the stone is thrown, the bird flies away in a horizontal straight line at a speed of 10m/s. Th stone reaches a heightdouble that of the pole and, in its descent, touches the bird. Find the horizontal component of the velocity of the stone.g=10m/s/s
\, and\, (x_2,h)\, \text{be the top of the poles respectively},\\ \text{V the velocity of the stone and (R,0) be the range}\, (\frac{R}{2},2h)\, \text{is the maximum of the parabola}\\ \text{I let}\, t=0,\, x=x_1,\, y=h\, \text{for the bird}\, \text{and}\, t=0,y=0,x=0\, \text{for the stone}.\, \text{Bird travels}\,(x_2-x_1)\, \text{in the same time it takes the stone to travel}\, x_2\\ \text{Parametric eqn for the stone is} (Vt\cos\alpha,Vt\sin\alpha-5t^2),\, \text{I have worked from here to get}\, V^2\sin^2\alpha=40h\\ \text{I know that the bird reaches}\, (x_2,h)\, \text{at the same time as the stone at}\, y=h\, \text{when}\, h=5t^2-Vt\sin\alpha\\ \text{From}\, 5t^2-Vt\sin\alpha+h=0\, \text{I know there are two solutions and I use the sum and products of roots to get}\\ t_1+t_2=\frac{V\sin\alpha}{5}\, \text{and}\, t_1t_2=\frac{h}{5}\, \text{to get}\, (t_2-t_2)^2=(t_1^2+t_2^2)^2-4t_1t_2=(\frac{V^2\sin^2\alpha}{25})-\frac{4h}{5}\\ \text{I worked all the way from here and I manage to get}\, V\cos\alpha=\sqrt{\frac{1500+50\sqrt{800}}{20}} )
I salvaged a lot of what you did. My method was mostly similar and I got the same answer
I don't think there's any real elegant way of doing this one. But this was my approach.
1. Noting that the maximum height of the stone is \(2h\), we can prove that \(\boxed{ h = \frac{V^2\sin^2\alpha}{40}} \). This gets used a lot.
2. Let \(T\) be the time of impact. The vertical equation of motion for the stone is \(y = V\sin \alpha - 5t^2\), and when \(t = T\), \( y = h\). The quadratic formula proves that \( \boxed{T = \frac{\sqrt2+1}{10\sqrt2}V\sin \alpha} \)
3. Let \(x_1\) be the x-coordinate of the pole and \(x_2\) be the x-coordinate of impact. The Cartesian equation of motion is \( y = -\frac{5x^2}{V^2}\sec^2\alpha + x\tan \alpha\). When \(y = h\), \(x = x_1\) or \(x = x_2\), so upon subbing \(y = h\), the sum and product of roots gives \( x_1 + x_2 = \frac{V^2\sin\alpha\cos\alpha}{5} \) and \(x_1x_2 = \frac{V^4\sin^2\alpha\cos^2\alpha}{400}\)
4. Using the identity \( (x_2-x_1)^2 = (x_2+x_1)^2 - 4x_1x_2 \) we can prove that \( \boxed{x_2 - x_1 = \frac{V^2\sin\alpha\cos\alpha}{5\sqrt2}} \)
5. I still have not considered the velocity of the bird. Since the bird's velocity is 10, its vertical equation of motion is \(x = 10t + x_1\). We know that when \(t = T\), \(x = x_2\). Subbing this in gives \( \boxed{T = \frac{x_2-x_1}{10}} \)
6. Subbing in everything, we ultimately get \( \boxed{V\cos \alpha = 5(\sqrt2+1)} \)
Edit: So perhaps the only improvement is that I didn't require \(t_1\); I just used \(x_1\). But I still needed \(t_2\); that's a bit inevitable. Details:
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