having recalled that \( \frac1{\cos^2\theta} = \sec^2\theta = 1+\tan^2\theta\).
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To understand what's going on here
There's two ways the ball can enter the hole. The ball can either be thrown at a very shallow angle and it just goes right through during its ascent (when it goes up).
OR, it can be thrown at a very steep angle, but still make its way through during its DESCENT instead (when it goes down)!
What we're actually given is basically the second case. We need to find the conditions for the first.
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Of course, since we have the Cartesian equation of motion, we can just directly use that instead of go back to the time equations.
having divided out by \(x\) since \(x=0\) is the initial starting point, which we're not interested in.
Explaining the bait
Recall that \(\boxed{m=\tan\theta} \).
The cause of the problem: What we're after is the range of the particle. Recall that the range of the particle is maximised when \(\theta = 45^\circ\). In other words, as you increase \(\theta\) from \(0^\circ\) up to \(45^\circ\), the range increases. But as you increase \(\theta\) from \(45^\circ\) to \(90^\circ\), the range decreases.
What I will do now is sketch \( R = \frac{40\tan\theta}{1+\tan^2\theta} \) on GeoGebra, with the four angles above included. (You can actually prove that in fact, \(R = 20\sin 2\theta\)).
(https://i.imgur.com/v4KbBQT.png)
The problem: If we just plug \(m=0.8\) and \(m=1.2\) blindly into \( \frac{40m}{1+m^2} \), we will miss out on the fact that there's a stationary point at \(45^\circ\). Plugging \( m = 0.8\) and \(m = 1.2\) will give our ranges \(R = 19.512\) and \(R = 19.672\). This would make us misbelieve that the width of the interval is just 19.672 - 19.512 = 0.160.
But in reality, when \(\theta=45^\circ\) and hence \(m=1\), we see \(R = 20\). This is actually a longer range! Hence the width of the interval is actually 20 - 19.512 = 0.488
Moral of the story: If you're given an interval, you need to consider any STATIONARY POINTS as well as the endpoints.
Remark: This problem does not occur for the other interval, since we don't bump into any more stationary points.
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In case you may have forgotten: \( \binom{n}{n-r} = \frac{n!}{(n-r)! [n-(n-r)]!} = \frac{n!}{r!(n-r)!} = \binom{n}{r} \)
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I write down explicitly how the cases work here in case you don't understand what comes next. Basically the next part just generalises all the cases, rather than does them one at a time. The method is the same.
This is the generic method: