Leaving Cert Applied Maths 2013

A ball is thrown vertically upwards with a speed of

$$44·1 m s ^{-1}$$       Calculate the time interval between the instants that the ball is 39·2 m above the  point of projection.  $$s = ut +\frac{1}{2}at^2$$ $$39.2 = 44.1t + \frac{1}{2}(-9.8)t^2$$ $$t^2 - 9t +8 = 0$$ $$(t-1)(t-8) = 0$$ ... $$t=1, t=8$$ $$t_1 = 8-1$$ $$t_1 = 7s$$

An aircraft P, flying at

$$600 kmh^{-1}$$ sets out to intercept a second aircraft Q, which is a distance away in a direction west 30 degrees south, and flying due east at $$600 kmh^{-1}$$ Find the direction in which P should fly in order to intercept Q.  $$\vec V_P = -600\cos a \vec i - 600\sin a \vec j$$ $$\vec V_Q = 600 \vec i$$ $$\vec V_{PQ} = \vec V_P - \vec V_Q = {-600\cos a - 600}\vec i - 600\sin a \vec j$$ $$\tan30 = \frac{600\sin a}{600\cos a+600}$$ $$\sqrt{3} \sin a = \cos a +1$$ $$3\sin^2 a = \cos^2 a +2\cos a +1$$ $$3(1 - \cos^2 a) = \cos^2 a + 2\cos a +1$$ $$0=4\cos^2 a + 2\cos a -2$$ $$\cos a = \frac{1}{2}$$ ... $$a = 60^{\circ}$$ ... $$W 60^{\circ} S$$ or $$S 30^{\circ} W$$

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