# 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
6 years 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).

- 6 years ago

Well, I believe that the gravitational field is fixed, so you may assume that acceleration is $a$....

Staff - 6 years 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.

- 6 years 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 - 6 years 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.

- 6 years ago

The above equation has two unknowns. Only one equation. Thus infinity solutions.

- 6 years ago

Yes , that was what I was saying too.

- 6 years ago

Hey guys, I've made this into a well-constrained problem, here

Staff - 6 years ago