A particle is projected at an angle $\theta$ above the horizontal where $\theta=\sin^{-1}\left(\dfrac{\sqrt{2}}{2}\right)$ . Find $\theta$ in degrees.

Model answer

Also accepted: 45 degrees, 45,0, 45.0

Explanation

\dfrac{\sqrt{2}}{2}=\dfrac{1}{\sqrt{2}}

, and the acute angle whose sine is

\dfrac{1}{\sqrt{2}}

45^{\circ}

. Its cosine is also

\dfrac{1}{\sqrt{2}}

, so the horizontal and vertical components of the launch velocity are equal.

Derived from ZIMSEC Mathematics 9164/4 Paper 4, June 2011, Q13

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