Water from a garden hose that is pointed?

Question

Water from a garden hose that is pointed 29° above the horizontal lands directly on a sunbather lying on the ground 4.1 m away in the horizontal direction. If the hose is held 1.4 m above the ground, at what speed does the water leave the nozzle?

Answer 1

The whole scenario is BS my friend...water will leave the hose at about 5 gpm if there was a sun bather in the nude even....ha ha

Answer 2

I'm not going to solve it but show you how.

Here's what you know. The water leaves the hose at 29 deg above horizontal.

The x distance traveled is 4.1 m

Vx (t) = 4.1 m

It leaves the hose 1.4 m above the ground so the y distance traveled is (-) 1.4 m.

Vyo (t) - (1/2)g(t^2) = -1.4 m

since the angle is 29 deg

Speed it leaves the nozzle is Vo

Vyo = Vo (sin (29))
Vx = Vo (cos (29)

So you have two equations and two unknows Velocity and time.
Should be able to solve it from here.

Answer 3

Let the nozzle velocity be v.
The vertical component of the nozzle velocity is v cos 29 deg.
Using this value, calculate the time it takes water leaving the nozzle to reach the ground:

Time to reach maximum height = (v cos 29 deg) / 9.8.
Vertical distance to max. height = 4.9 (time to max height)^2
Max height = vert. dist. to max. height + 1.4 (which was the nozzle height)
Time to fall from max height to ground = sqrt (2 max. ht. / 9.8)
Total time = Time to reach max. ht. + Time to fall from max. ht. to ground

Horizontal distance = Total time x Horizontal component of nozzle velocity = Total time (v cos 29 deg) = 4.1 m (as given in the problem.

So you need to write out all the above algebra using v as an unknown variable. You will end up with one equation in one unknown (v), which you can solve for v.