Answer to Question #124387 in Mechanics | Relativity for Abhi

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

Explanations & Calculations

• Refer to the figure above.

• The screen should be moved from point A to point B in order for to fulfill the given situation & hence the parabolic path is symmetrical about the center axis.
• This problem could be solved considering parabolic behavior or projectile motion.
• Projectile motion is somewhat lengthier compared to parabolic method.

• Consider the parabolic relationship — $\bold{y} = \bold{ax^2+bx+c}$ — for an arbitrary point P(x,y) on the curve.
• By differentiating y with respect to x give the gradient/slope of the function at a selected point.
• Therefore, by performing this at the origin [(0,0) point or at the start/projection] it's possible to find the projection angle.

• Therefore,

$\qquad\qquad \begin{aligned} \small \frac{dy}{dx} &= \small 2ax+b\\ \small \frac{dy}{dx}_{(0,0)}&= \small tan\theta = 2a(0) + b \\ \small \frac{dy}{dx}_{(0,0)} &=\small b \end{aligned}$

the value of b should be found.

1) Considering the coordinates of origin,

$\qquad\qquad \begin{aligned} \small 0m &= a(0^2) +b(0) + c\\ \small c &= \small 0 \end{aligned}$

2) For point A,

$\qquad\qquad \begin{aligned} \small 4m &= \small a(6^2)+b(6)\\ \small 2&= \small 18a +3b\\ \small a &= \small \frac{2-3b}{18} \cdots\cdots(1) \end{aligned}$

3) For point B,

$\qquad\qquad \begin{aligned} \small 4m &= \small a(10^2)+b(10)\\ \small 2&= \small 50a+5b\\ \small a &= \small \frac{2-5b}{50} \cdots\cdots(2) \end{aligned}$

4) By (1) = (2) we get,

$\qquad\qquad \begin{aligned} \small \frac{2-3b}{18} &= \small \frac{2-5b}{50}\\ \small b &= \small 1.0667 \end{aligned}$

5) Then,

$\qquad\qquad \begin{aligned} \small \theta &= \small tan^{-1}(1.0667)\\ \small &= \small 46.8476^0 \end{aligned}$

• Therefore, projection angle is = $\theta = \bold{46.85^0}$