Question

A container of water, diameter 12 cm, has a small opening near the bottom that can be unplugged so that the water can run out. If the top of the tank is open to the atmosphere, what is the exit speed of the water leaving through the hole. The water level is 15 cm above the bottom of the container. The center of the 3.0 diameter hole is 4.0 cm from the bottom. How long does it take the water to hit the ground? How far from the container will the water land?

Answer 1

Answer:

1.32 m

Explanation:

from the given figure we can find the velocity at the hole in the container

let it be v and we know that

then $v= \sqrt{2gh}$

h= 15-4= 11 cm , g=9.81

$v= \sqrt{2(9.81)(11)}$

v=14.69 m/s

now using

s=ut+0.5at^2

t= is the time required by water to reach bottom of the container

u = velocity in vertical direction is zero.

therefore,

$0.04= 0\times t+\frac{1}{2}\times9.81\times t^2$

t= 0.09030 sec

let x be the distance far from the container will the water land

x=vt

x=14.69×0.09030 = 1.32 m