Q.(a) Assume that water issuing from the end of a horizontal pipe, 7.5 m above the ground, describes a parabolic path. The vertex of the parabolic path is at the end of the pipe. At a position 2.5 m below the line of the pipe, the flow of water has curved outward 3 m beyond the vertical line through the end of the pipe. How far beyond this vertical line will the water strike the ground? OR

(a) models the water jet as a parabola with vertex at the pipe end, fits it using the one given point, then finds where it meets the ground 7.5 m below; (b) proves the cosine addition formula using the dot product of two unit vectors.

(a) Parabolic path of the water jet

1. Set up axes with the vertex (pipe end) at the origin: $x$ measured horizontally outward, $y$ measured vertically downward (the direction the water falls). Since the vertex is at the origin and the path opens in the $+y$ (downward) direction, its equation has the form $x^2=4ay$.
2. Given data: at $y=2.5$ m below the pipe, the water has moved $x=3$ m outward. Substitute: $3^2=4a(2.5)\Rightarrow 9=10a\Rightarrow a=0.9$.
3. So the path is $x^2=4(0.9)y=3.6y$.
4. The ground is $7.5$ m below the pipe, i.e. $y=7.5$. Substitute: $x^2=3.6\times7.5=27\Rightarrow x=\sqrt{27}=3\sqrt3$.
5. Numerically $3\sqrt3\approx5.196$ m $\approx5.2$ m.

(b) Vector proof of $\cos(\alpha+\beta)$

1. Let $\hat a$ be the unit vector along $OA$ making angle $\alpha$ with the positive $x$-axis (measured above it), and $\hat b$ the unit vector along $OB$ making angle $\beta$ with the positive $x$-axis but measured below it, so the angle between $\hat a$ and $\hat b$ is $\alpha+\beta$.
2. In components: $\hat a=\cos\alpha\,\hat i+\sin\alpha\,\hat j$, $\hat b=\cos\beta\,\hat i-\sin\beta\,\hat j$. …