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)
A monkey escapes from the zoo and climbs a tree. After failing to entice the monkey down, the zookeeper fires a tranquilizer dart directly at the monkey. The monkey lets go at the instant the dart leaves the gun. Show that the dart will always hit the monkey, provided that the dart reaches the monkey before he hits the ground and runs away.
\(S_{monkey} = (1/2)gt^2\),
so \(h_{monkey} = h-(1/2)gt^2\)
\(d = vcos(\theta)t\),
so \(dtan(\theta) = vtsin(\theta) = h\)
\(h_{bullet} = S_{bullet} = vsin(\theta)t-(1/2)gt^2 = h-(1/2)gt^2 = h_{monkey}\),
so the bullet will always hit the monkey
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