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?
You are watering a garden. The height
4 Answers
I think the problem is something wrong.
All is okk but finally the value of
Explanation:
Please help me further if I am wrong.
It is impossible to increase the range by 4 m.
Explanation:
Given,
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
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
Now
Comparing this with the general equation for a parabolic trajectory
which yields
we get
The largest range possible at this speed is
See below.
Explanation:
The parametric movement kinematic equations are
and the path equation is obtained by time elimination.
Calling
Regarding the coordinates
at
This range is maximum at
In the present case
Solving for
so the maximum possible reach is
Now if the actual reach is
we conclude that the proposed elongation concerning the watering range is unattainable.
Let
#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
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
#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
.-.-.-.-.-.-.-.-.-.-.-.-.--.
Modeled equation graph without shifting of origin, as in the first figure above.
Left part of above graph expanded to show negative root.