- #1

Insolite

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**Question:**

A cannon, located ##60.0 m## from the base of a vertical ##25.0 m## tall cliff, shoots a ##15 kg## shell at ##43.0°## above the horizontal toward the cliff. What must the minimum muzzle velocity be for the shell to clear the top of the cliff?

I have attempted a solution that I'm doubtful is correct as I have made the assumption that the shell glances the edge of the cliff at the highest point of its trajectory and I'm not certain whether this is necessary (as in it may just clear the cliff at a point in its trajectory that is before the highest point). However, without making this assumption I don't know how to proceed with the question.

**Solution:**

Finding the initial ##y##-component of velocity, where the max height of projectile is assumed to be the height of the cliff, ##25 m##, where ##v_{y} = 0## -

##v_{y}^{2} = v_{0y}^{2}sin^{2}{α} - 2g(25)##

##0 = v_{0y}^{2}sin^{2}{43°} - 2g(25)##

##50g = v_{0y}^{2}sin^{2}{43°}##

##v_{0y} ≈ 32.47 ms^{-1}##

Finding the initial ##x##-component of velocity, where I have reversed the set-up and calculated the time taken for the projectile to go from ##v_{0y} = 0## to ##v_{y} = 32.47 ms^{-1}## in the ##y##-axis -

##v_{y} = v_{0y} - (-g)t##

##v_{y} = gt##

##t ≈ \frac{32.47}{9.81}##

##t ≈ 3.31 s##

Finding the initial ##x##-component of velocity -

##x = (v_{0x}cosα)t##

##v_{0x} = \frac{x}{tcos43°}##

##v_{0x} = \frac{60.0}{3.31cos43°}##

##v_{0x} ≈ 24.78 ms^{-1}##

The magnitude of the initial muzzle velocity -

##v = \sqrt{v_{0x}^{2} + v_{0y}^{2}}##

##v = \sqrt{24.78^{2} + 32.47^{2}}##

##v = 40.9 ms^{-1}##

Any help or advise is appreciated.

Thanks.