Question

In designing a backyard water fountain, a gardener wants to stream of water to exit from the bottom of one tub and land in a second one. The top of the second tub is 0.5 m below the hole in the first tub, which has water in it to a depth of 0.15 m. How far to the right of the first tub must the second one be placed to catch the stream of water?

Answer 1

To solve this problem it is necessary to apply the concepts related to the kinematic equations of movement description.

From the definition we know that the speed of a body can be described as a function of gravity and height

$V = \sqrt{2gh}$

$V = \sqrt{2*9.8*0.15}$

$V = 1.714m/s$

Then applying the kinematic equation of displacement, the height can be written as

$H = \frac{1}{2}gt^2$

Re-arrange to find t,

$t = \sqrt{2\frac{h}{g}}$

$t = \sqrt{2\frac{0.5}{9.8}}$

$t = 0.3194s$

Thus the calculation of the displacement would be subject to

$x = vt$

$x =1.714*0.3194$

$x = 0.547m$

Therefore the required distance must be 0.547m