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Projectile practice

You have a cannon a distance \(d\) before the start of a hill. The hill goes up at an angle \(\theta\) from the ground. What angle and velocity must you shoot a cannon ball so that it lands a distance \(s\) up the hill?(The distance s is measured from the bottom of the hill, i.e. the distance along the hill).

Note by Nihar Mahajan
1 year, 4 months ago

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I think there are infinite solutions to this problem. There are infinite parabolas passing through (0,0) and \((d+sCos\theta,sSin\theta)\), the cannon at (0,0). Niranjan Khanderia · 1 year, 4 months ago

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@Niranjan Khanderia I got \(\large v= (d+s\cos\theta)\sqrt{\dfrac{g}{2(\sin 2\alpha)(d+s\cos\theta) - s\sin\theta\cos^2\alpha}}\) where \(v\) and \(\alpha\) is the required velocity and angle. As you say , there are infinitely many solutions. Is this correct?

This question is given to me by @Josh Silverman Sir. Nihar Mahajan · 1 year, 4 months ago

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@Nihar Mahajan The above equation has two unknowns. Only one equation. Thus infinity solutions. Niranjan Khanderia · 1 year, 4 months ago

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@Niranjan Khanderia Yes , that was what I was saying too. Nihar Mahajan · 1 year, 4 months ago

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@Niranjan Khanderia Well, I believe that the gravitational field is fixed, so you may assume that acceleration is \(a\).... Calvin Lin Staff · 1 year, 4 months ago

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@Calvin Lin Niranjan sir is right since angle and velocity both are upto us there are infinite possibilities.One of them has to be given for a finite(2 if velocity is given[not always]) number of solutions. Krishna Sharma · 1 year, 4 months ago

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@Krishna Sharma Oh yes that's true. For some reason, I though the angle of the cannon was fixed (when it's just the angle of the hill that is fixed).

So, what is the relationship between \( \theta \) and \( v \)? Calvin Lin Staff · 1 year, 4 months ago

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Hey guys, I've made this into a well-constrained problem, here Josh Silverman Staff · 1 year, 4 months ago

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