A particle is projected from the point O with speed 8 ms $^{-1}$ at an angle $\theta$ above the horizontal, and it passes through the point $A(8;\ -1.81)$ , distances in metres, with O as origin. Taking $g=9.81$ ms $^{-2}$ , there are two possible values of $\theta$ . Find the smaller one, in degrees, correct to 1 decimal place.

Model answer

32,2

Also accepted: 32.2, 32, 32,2°, 32.2°, 32,24, 32.24

Explanation

Using

y=x\tan\theta-\dfrac{gx^{2}}{2u^{2}}\left(1+\tan^{2}\theta\right)

with

x=8

y=-1.81

u=8

and

g=9.81

-1.81=8\tan\theta-4.905\left(1+\tan^{2}\theta\right)

, so

4.905\tan^{2}\theta-8\tan\theta+3.095=0

and

\tan\theta=\dfrac{8\pm\sqrt{3.2761}}{9.81}=\dfrac{8\pm1.81}{9.81}

. That gives

\tan\theta=1

\tan\theta=0.631

, so

\theta=45^{\circ}

\theta=32.2^{\circ}

Derived from ZIMSEC Mathematics 9164/4 Paper 4, November 2004, Q14

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