I have a question involving kinematics and physics in a plane. The question is as follows:

I am working with the two fundamental kinematics equations:

Code (Text):

x = v_ix t

y = v_iy t - 1/2 gt^2

(An underscore, _, indicates subscripts. A caret, ^, indicates superscripts.)

Since the final positions of both the receiver and the ball must be identical, I have tried endlessly to modify equations which both involve position (that is, isolating x and then making one formula equal to another). However, some other variable (usually time) always gets in the way of my solving the equation. Any hints which would set me on the right track are appreciated.

As with most of these problems, the confusion arises because the student does not develop a sufficient notation (or he develops an unclear notation, which is not the case here).

Relative to Fred's position, Doug has the position as a function of time:
[tex]x_{D}(t)=6t+20[/tex]
Again, relative to Fred, the ball has the coordinate positions:
[tex]x_{B}(t)=v_{x}t=v_{0}\cos(\theta_{0})t, y_{B}(t)=v_{y}t-\frac{gt^{2}}{2}=v_{0}\sin(\theta_{0})t-\frac{gt^{2}}{2}[/tex]
where [itex]v_{0}[/itex] is the initial speed you are to find, and [itex]\theta_{0}[/itex] is the initial angle (given as 40 degrees).

Thus, essentially, you are to solve the system of equations for [itex]v_{0}[/itex], the other unknown being the time t:
[tex]6t+20=v_{0}\sin(\theta_{0})t[/tex]
[tex]v_{0}\sin(\theta_{0})t-\frac{gt^{2}}{2}=0[/tex]
knowing that [itex]t\neq{0}, v_{0}>0[/itex]
Agreed?