Question

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?

Answer 1

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

Explanation:

Please help me further if I am wrong.

Answer 2

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.

Answer 3

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.

Answer 4

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^@`

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`

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.

Left part of above graph expanded to show negative root.