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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, 11 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). · 1 year, 11 months ago

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. · 1 year, 11 months ago

The above equation has two unknowns. Only one equation. Thus infinity solutions. · 1 year, 11 months ago

Yes , that was what I was saying too. · 1 year, 11 months ago

Well, I believe that the gravitational field is fixed, so you may assume that acceleration is $$a$$.... Staff · 1 year, 11 months ago

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. · 1 year, 11 months ago

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 $$v$$? Staff · 1 year, 11 months ago