You posted a question about having two cannon balls strike the same target simultaneously, with a time delay between shots:

Question

https://www.jiskha.com/display.cgi?id=1520556373

I have had some time to think about it, and have come up with the following ideas. Maybe you have solved it in the meantime.

Since the range of a shot fired with velocity v at angle θ is
R(θ) = v^2/g sin2θ
then if the 2nd shot, fired at angle Ø with the same speed is the same, we need
sin2θ = sin2Ø
That means that 2θ = 180-2Ø, or Ø=90-θ

Our cannonball was fired with v=1000, and the second shot was delayed k seconds.
Since the time in the air at angle θ is 2v/g sinθ = 200sinθ, we need to maximize the distance traveled by the ball. That is, maximize the flight time. This is clearly when θ=90 degrees.

But, that means the ball goes straight up, making it impossible to fire a 2nd ball at a lower angle and hit the same spot. So, pi/4 < θ < pi/2.

So, with a delay of k seconds, we have

200sinθ = 200sinØ + k
k = 200(sinθ-sinØ)
But since Ø=90-θ, sinØ=cosθ and so
k = 200(sinθ-cosθ)

Sanity check: If the time delay is zero, then sinθ=cosθ and θ=45. This provides the maximum range, so no time delay can be allowed. But, we need k>=60

Since we need to wait at least a minute, k >= 60, meaning that
200(sinθ-cosθ) > 60
sinθ-cosθ >= 0.3

Now we know that we need to maximize θ subject to the constraint that
pi/4 < θ
sinθ-cosθ > 0.3

This means that 0.999154 < θ < pi/2
In other words, we basically need θ=1

check:
Range(θ=1) = 90929 -- k=60.24
Range(θ=1.1) = 80849 -- k=87.52
Range(θ=0.9) = 97384 but the time lag is only 32.37 seconds

So, you can see that larger values of θ result in a smaller range, and smaller θ means that you cannot wait the required 60 seconds before firing the 2nd shot.