Watering a garden?

You are watering a garden. The height h (in meters) of water spraying from the garden hose can be modeled by h =-0.1x^2 + 1.4x +1, where x is the horizontal distance (in meters) from where you are standing. At the same point you change the hose angle so that the water hits the ground 4 [m] farther from the initial hit point. What angle is required to get the last situation?

4 Answers
Feb 25, 2018

I think the problem is something wrong.
All is okk but finally the value of sintheta' comes wrong.

Explanation:

my notebook...my notebook...

Please help me further if I am wrong.

Feb 25, 2018

It is impossible to increase the range by 4 m.

Explanation:

Given, h=-0.1x^2+1.4x+1, we get, for h=0:
-x^2+14x+10=0. This has two roots

x = {-14pmsqrt{14^2-4times(-1)times 10}}/(2times (-1))=7pm sqrt{59}

The two roots correspond to the two points at which the spray is at ground level - once, where it starts from the hose - the other, where it hits the ground. The horizontal range for the spray is thus 2sqrt{59}.

It is impossible to say from the trajectory which root corresponds to the starting point (there is no information about the direction in which the spray is traveling) but this will not affect the final answer. For the sake of definiteness I consider x_0 = 7-sqrt{59} as the starting point.

Now
{dh}/dx = -0.2x+1.4,qquad {d^2h}/dx^2=-0.2

Comparing this with the general equation for a parabolic trajectory

h = (x-x_0)tan alpha -{g(x-x_0)^2}/{2u^2 cos^2 alpha}

which yields

{dh}/dx = tan alpha -{g(x-x_0)}/{u^2cos^2 alpha}, quad {d^2h}/dx^2 = -g/{u^2cos^2 alpha}

we get

tan alpha = {dh}/dx|_{x=x_0} = -0.2(7-sqrt{59})+1.4=1.536
g/{u^2 cos^2 alpha}= 0.2 implies u^2/g = sec^2 alpha/0.2={1+tan^2 alpha}/0.2=16.8

The largest range possible at this speed is u^2/g = 16.8 m. The problem, however, asks for the the range to be increased from the initial 2*sqrt{59}~~15.36 m by 4 m to 19.36 m. This is impossible at the given speed.

Feb 25, 2018

See below.

Explanation:

The parametric movement kinematic equations are

x = x_0 + (v_0 cos theta) t
y=y_0 +(v_0 sin theta) t-1/2g t^2

and the path equation is obtained by time elimination.

Calling delta x = x-x_0
y-y_0 = delta y = (v_0 sin theta)/(v_0 cos theta)delta x-1/2 g((delta x)/(v_0 cos theta))^2

Regarding the coordinates deltax, delta y the path hits the ground at

delta y = 0 = ((v_0 sin theta)/(v_0 cos theta)-1/2 g(1/(v_0 cos theta))^2delta x)delta x

at delta x = 0 (origin) and at delta x = (2v_0^2)/g sin theta cos theta (hit point)

This range is maximum at theta = pi/4 as can easily be verified. Then

delta x_(max) =( 2v_0^2)/g1/2 = v_0^2/g

In the present case

y = -0.1x^2+1.4 x+1 and by comparing paths,

(dy/dx)_(x=0) = tan theta_0=1.4
(g/2)1/(v_0^2cos^2theta_0)=0.1

Solving for theta_0,v_0^2 gives

v_0^2 = 14.8 g

so the maximum possible reach is

deltax_(max) = 14.8

Now if the actual reach is deltax_a=14 (Here deltay_a = y-1)

we conclude that the proposed elongation concerning the watering range is unattainable.

Feb 26, 2018

Let h_1=h-1. Our equation becomes

h_1=-0.1x^2+1.4x
h_1^'=-0.2x+1.4
h_1^'=1.4=0.136
theta=54.46^@

enter image source here

Trajectory formula is given as

h_1=xtan theta-g/(2v_0^2cos^2theta)x^2

Comparing with given expression we have

g/(2v_0^2cos^2theta)=0.1
=>v_0^2=g/(2xx0.1cos^2theta)
=>v_0^2=14.8g
=>v_0=12.0\ ms^-1

Max Range is for angle of projection theta=45^@.
New time of flight for this angle can be found from

s=ut+1/2at^2

Reverting back to origin at the location of hose.

-1=12.0\ sin45^@t-1/2xx9.8t^2
=>4.9t^2-6sqrt2t-1=0

Ignoring the first solution, found using in-built graphics utility, as time can not be -ve. Valid solution is

t=1.842\ s

enter image source here

New range is

R_max=(v_0costheta) t
R_max=(12.0\ cos45^@) 1.842
R_max=15.63\ m

Hence, hitting at a distance of 4\ m farther from the initial hit-point at 14.68\ m as shown in the figure below is not possible.

.-.-.-.-.-.-.-.-.-.-.-.-.--.
Modeled equation graph without shifting of origin, as in the first figure above.
enter image source here

Left part of above graph expanded to show negative root.

enter image source here