A projectile is thrown from a point P. It moves in such a way that its distance from P is always increasing. Find the maximum angle above the horizontal with which the projectile could have been thrown. You can ignore air resistance.
Distance, \(D = \sqrt{x^2+y^2}\)
\(D^2 = x^2+y^2\)
For distance from P always increasing, \(\frac{dD}{dt} > 0\)
(To get an easier calculation, we use \(\frac{d(D^2)}{dt} > 0\))
\(D^2 = (u \cos\theta \cdot t)^2 + (u \sin\theta \cdot t - \frac{1}{2}gt^2)^2\)
\(= u^2\cos^2\theta \cdot t^2 + u^2\sin^2\theta \cdot t^2 + \frac{1}{4}g^2t^4 - u\sin\theta gt^3\)
\(= u^2 t^2 - u\sin\theta gt^3 + \frac{1}{4}g^2t^4\)
\(\frac{d(D^2)}{dt} = 2u^2t - 3u\sin\theta gt^2 + g^2t^3\)
\(= t(2u^2 - 3u\sin\theta g t + g^2t^2) > 0\)
For \(t>0\),
So, \(2u^2 - 3u\sin\theta g t + g^2t^2 \ge 0\)
\(g^2t^2 - 3u\sin\theta g t + 2u^2 \ge 0\)
\(\Delta = (3u\sin\theta g)^2 - 4 \cdot g^2 \cdot 2u^2\)
\(= 9u^2\sin^2\theta g^2 - 8g^2u^2\)
\(= u^2g^2(9\sin^2\theta - 8) < 0\)
For \(u^2g^2 > 0\)
So, \(9\sin^2\theta - 8 < 0\)
\(\sin^2\theta < \frac{8}{9}\)
\(\sin\theta < \frac{2\sqrt{2}}{3}\)
So, \(\theta_{max} = \arcsin\left(\frac{2\sqrt{2}}{3}\right) \approx 70.5^{\circ}\)
(no real roots)
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