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## Homework Statement

Hi,

I am doing an end of the year calculus presentation on projectile motion in human cannonballs. As an introduction to the projectile motion equations, my group found this problem:

I'm not a physics student, so thank you for any help!

I am doing an end of the year calculus presentation on projectile motion in human cannonballs. As an introduction to the projectile motion equations, my group found this problem:

The original problem with a figure of it can be found http://higheredbcs.wiley.com/legacy/college/anton/0471482730/calc_horizons/blammo.pdf" [Broken]Blammo is to be fired from 5mabove ground level with a muzzle velocity of 35m/s over a flaming wall that is 20 m high and past a 5-m-high shark pool. To make the feat impressive, the pool will be made as long as possible. Your job as Blammo’s manager is to determine the length of the pool, how far to place the cannon from the wall, and what elevation angle to use to ensure that Blammo clears the pool.

I'm not a physics student, so thank you for any help!

## Homework Equations

I'm sorry, but I couldn't figure out the latex equations...Hope these are clear enough.

Projectile Motion Formulas

R=(v

r(t)=v

(Are there equations I'm missing for this problem that would be more useful/better than these??)

Projectile Motion Formulas

R=(v

_{o}^{2}/g)sin(2q)r(t)=v

_{0x}t**i**+(v_{oy}t-.5gt^{2}**j**which, if I'm correct (??) breaks down into:

x(t)=x

y(t)=y

x(t)=x

_{o}+v_{ox}ty(t)=y

_{o}+v_{oy}t-.5gt^{2}(Are there equations I'm missing for this problem that would be more useful/better than these??)

## The Attempt at a Solution

Okay...Here we go:

R=(v

x(t)=x

y(t)=y

R+x

R=(v

_{o}^{2}/g)sin(2q)For the range of the cannon (R) to be maximum, the angle of elevation (q) must equal 45

R=(35

R=125.026m

^{o}.R=(35

^{2}/9.798)sin(2*45)R=125.026m

x(t)=x

_{o}+v_{ox}ty(t)=y

_{o}+v_{oy}t-.5gt^{2}@45

v

v

I placed the wall of fire at x=0. So, solving for x

0=x

t=-x

Sub t into the y(t) equation:

y(t)=20=5+v

15=-x

x

^{o}, v_{ox}=v_{oy}.v

_{ox}=v_{o}cos(q)v

_{ox}=v_{oy}=24.749m/sI placed the wall of fire at x=0. So, solving for x

_{o}would give the distance from the wall the cannon must be placed. (Correct??)0=x

_{o}+v_{ox}tt=-x

_{o}/v_{ox}Sub t into the y(t) equation:

y(t)=20=5+v

_{oy}t-.5gt^{2}15=-x

_{o}-4.899(-x_{o}/24.749)^{2}x

_{o}=-17.43m

R+x

_{o}=length of poolThe pool is 107.596m long.

So, I think I have solved this correctly. One thing I'm worried about is the assumption of q=45

^{o}. I know this makes R the greatest, but does it make the length of the pool the greatest? If the angle was steeper, could the cannon be placed closer to the fire, making the pool be a greater part of the path?

Any help or confirmation of my answers would be

**greatly**appreciated. Thank you so much!!

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