The figure shows a stream of water flowing through a hole at depth h = 9.87 cm i

Question

The figure shows a stream of water flowing through a hole at depthh= 9.87 cm in a tank holding water to heightH= 39.1 cm.(a)At what distancexdoes the stream strike the floor?(b)At what depth should a second hole be made to give the same value ofx?(c)At what depth should a hole be made to maximizex?

Give your answers in cm.

Chapter 14, Problem 71

Explanation / Answer

(a) Set up Bernoulli's equation along a streamline from the liquid surface (') to the point directly after the outlet: p'/? + g·H + v'²/2 = p/? + g·h + v²/2 The pressure at the two point is equal to the ambient pressure. p'=p (as long the tank is open to atmosphere or has vent) Furthermore the motion of liquid surface in the tank is small compared to the motion at the outlet. Therefore you neglect it: v' ˜ 0 Hence g·H =g·h + v²/2 Solve for the outlet velocity: v = v(2·g·(H-h)) (The fact that under this assumptions the outlet velocity is proportional to height of liquid level above outlet is also known as Torricelli's law) Because you can neglect viscous effects, you can treat the liquid drops a projectile moving through the air. The range of such a projectile fired horizontally at height h is: (for derivation see 3rd link set d=x y0 = h and ? = 0°): x = v·v(2·h/g) With the outlet velocity from above you find: x = v(2·g·(H-h))·v(2·h/g) = 2·v((H-h)·h) = 2·v(264)cm ˜ 32.5 cm (b) For every distance there are two possible height: x = 2·v((H-h)·h) If know one value h leading to a distance x, the other leading to the same distance is: h' = H-h => h' = 34cm -12 cm = 22cm (c) x = 2·v(H·h-h²) At the maximum the first derivative is equal to zero: dx/dh = 0 (H - 2·h) /v(H·h-h²) = 0 (H - 2·h) = 0 h = H/2 = 17cm For hH/2 it is postive. Hence we have change from positive to negative slope at h=H/2, that means there is maximum at this point.

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