(Early college) Confused about a system of equations?

[kylemm]

Not sure if this falls under algebra, trig or what but I'm stuck on how to solve this problem in an informal homework assignment:

Suppose a ball moves horizontally at a constant velocity v. Meanwhile, a second ball moves with constant acceleration a along a line oriented at an angle theta from the vertical. At time 0, the first ball is a distance h directly above the second ball, and the second ball is instantaneously at rest.
What angle theta would result in a collision at some later time t? The answer should involve the known values v, h, and a (Let us assume that v, h, and a are all positive).
This collision implies that the positions (x and y coordinates) of both balls are the same at time t. This gives us two equations, one for the x coordinate and one for the y coordinate:
(1/2)at²sin(theta) = v*t
h = (1/2)at²cos(theta)


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Now, I don't really understand all of the specifics of this problem as a first year student who couldn't take physics in high school. And this is a 1 credit online course so it's difficult to find someone to ask. But I tried solving it as a system of equations with t and theta as our variables of interest, and ended up with t = (2v)/(a*sin(theta)), plugged that in to the second equation and got theta = arccos(1-((2v²)/(ah))) but that's apparently not right. Any help with how to approach this problem would be greatly appreciated!

 

[Ishuda]

Suppose a ball moves horizontally at a constant velocity v: This says that y (height) is constant and, without loss of generality we can take that height to be h (At time 0, the first ball is a distance h directly above the second ball). It also says that x moves the same distance each time slice and, again without loss of generality, we can take the starting point at x=0. So
x1 = v * t
y1 = h

A second ball moves with constant acceleration a along a line oriented at an angle theta from the vertical. At time 0, the first ball is a distance h directly above the second ball, and the second ball is instantaneously at rest. So the distance traveled, s, by the second ball is
s = d t² + b t + c
where a is the acceleration and d = a/2, b is the speed at time zero and c is the initial distance at which it starts or
s = d t²
since the initial velocity is zero (At time 0, the second ball is instantaneously at rest) and the initial height is zero (see the first ball's position at t=0). Since the direction must be up for the balls to collide and the angle is measured from the vertical,
x2 = s sin(\(\displaystyle \theta\))
y2 = s cos(\(\displaystyle \theta\))

If the balls are going to collide at time t then, at time t, x2=x1 and y2=y1 or
d t² sin(\(\displaystyle \theta\)) = v t => t = \(\displaystyle \frac{v}{d sin(\theta)}\) (which is what you got)
and thus
\(\displaystyle d (\frac{v}{d sin(\theta)})^2 cos(\theta)=h\) (which, somewhere in here, is where you made your mistake)
or, letting u = \(\displaystyle cos(\theta)\)
\(\displaystyle d h u^2 + v^2 u - d h = 0\)

Use the quadratic formula to find u [the negative square root can be discarded for physical reasons], thus \(\displaystyle cos(\theta)\) and \(\displaystyle \theta\), and then t which will lead to the point of collision.