Well the answer is about 5.5° and let's see how we calculate that.
Well first, we need to calculate the range of the cannon. How do we do that?
Let's look at our initial velocity vector.
The cannon fires at the speed of 56 m/s, an angle of 45° because that optimizes the range.
Now we want to break this into components--V₀x and V₀y.
We can do some trigonometry to show that the x and y components are both equal to 39.6 m/s.
If I'm going to know how far this cannonball is going to go, I should
probably know how long it's going to spend in the air.
Let's use the y component velocity to do that.
Looking at the trajectory of this ball, I want to first find out how long it takes to get to the very top
that's when the vertical component of velocity would be equal to zero.
Using this equation and plugging in what we know, and remember in this case acceleration
is downward so negative, we find it takes 3.96 seconds to get to the top of the trajectory,
which means it would take another 3.96 to get down which results in a total time of 7.92 seconds.
Now we known the x direction, speed isn't going to be changing.
There is no acceleration in the x direction.
So if I look at this equation, this term goes to 0 and we're left with this.
So what's the range of the cannon, how far can it reach Δx or it's just
the x component of velocity 39.6 times the time, which gives us 314 m.
If you got this far, good work. We're almost towards the end.
Now we just need to convert this somehow into an angular size.
Well looking at this angle β, we know this flagpole has a height of 30 m
and in fact, it looks like we know the opposite and the adjacent sides.
So we can write the tangent of β is 30/314.
If I ask Google what the arctan of 30/314, in degrees because remember that's how we go
from ratios to angles, it gives me a value of 5.5°--good work!