(c14p55) A tank is filled with water to a height H, 40 m . A hole is punched in

Question

(c14p55) A tank is filled with water to a height H, 40 m. A hole is punched in one of the walls at a depth h, 9.20 m, below the water surface (see the figure). What is the distance x from the base of the tank to the point at which the resulting stream strikes the floor?

Could a hole be punched at another depth to produce a second stream that would have the same range? If so, at what depth?

At what depth should a hole be punched to make the emerging stream strike the ground at the maximum distance from the base of the tank?

Explanation / Answer

H = 40 m, h =9.2 m

voy =0
ay =g

From kinematic equation along y direction

S =voy+(1/2)ayt^2

H -h 0+(1/2)gt^2

t = [2(H-h)/g]^1/2

vx = [2gh]^1/2

The horizontal distance covered in this time inerval

x = vxt = [2gh]1/2[2(H-h)/g]1/2

x =2[(H-h)h]1/2

x =2[(40-9.2)*9.2)]1/2

x = 33.67 m

(b) Yes

x =2[(H-h)h]1/2

x2 = 4Hh - 4h2

33.67*33.67 = (4*40*h)- 4h2

h2 - 40h +283.42 =0

by solving above equation we get h =9.2 m

h = 30.8 m

(c) x =2[(H-h)h]1/2

deferentiating withrespect to h

dx/dt = (H-2h)/(Hh-h2)1/2

To get maximum disance from base of tank dx/dt =0

0 = = (H-2h)/(Hh-h2)1/2

By solving this we get

h =H/2 =40/2 =20 m

maximum disance from base of tank is =20m

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