Question

A jet of water, discharged from a nozzle, hits a screen 6 m away at a height of 4 m above
the centre of a nozzle. When the screen is moved 4 m further away, the jet hits it again at
the same point. Assuming the curve described by the jet to be parabolic, find the angle at
which the jet is projected. (Ans. 46° 51')

Answer 1

Assume that $y=ax^2+bx+c$

If $x=0$ then $y=0$ .We have $0=a\cdot0+b\cdot0+c\to c=0$.

For point (6,4): $4=a\cdot6^2+b\cdot6\to a=\frac{4-6b}{36}$

For point (10,4): $4=a\cdot10^2+b\cdot10\to a=\frac{4-10b}{100}$

So,

$\frac{4-6b}{36}=\frac{4-10b}{100}\to b=1.067$

At point (0,0)

$\frac{dy}{dx}=2ax+1.067=2a\cdot0+1.067=\tan(\alpha)$

$1.067=\tan(\alpha)\to \alpha=46,85°$