Question #124391

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')

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')

Expert's answer

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\u00b0"

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