### The Projectile Incident (updated)

Craig Miller sent me an e-mail:*Since you gave numbers it would be a shame not to try and figure out the answer. So doing some simplified calculations that ignore factors like drag, spin, etc. and go on pure speed, distance, and time here are the calculations I start with:*

*Variables*

*V(x)= horizontal velocity ft/s at mower*

*V(y)= vertical velocity ft/s at mower*

*V(t)= total velocity*

*T(1)= Time when it hits the door s*

*T(2)= Time when it hits the wall s*

*Equations*

*T(1)*V(x)=25ft*

*T(1)*V(y)-1/2(T(1)^2*32.17ft/s^2)=2.166ft*

*T(2)*V(x)=40ft*

*T(2)*V(y)-1/2(T(2)^2*32.17ft/s^2)=1.75ft*

*A whole bunch of math, conversions and stuff goes here.*

*This yields*

*V(x)=74.978ft/s*

*V(y)=11.861ft/s*

*V(t)=75.910ft/s*

*T(1)=.333s*

*T(2)=.533s*

*Yielding a rock speed of roughly 51.75mph as it leaves the mower.*

*Now these numbers are probably off, since there are several assumptions here, such as the rock having a vertical height of 0 at T(0). Also things like lawn elevation could potentially have huge impacts on the resulting equations.*

*In reality the speed would be higher, simply because the window would take energy out of the rock, therefore slowing both it's horizontal and vertical speed, but that would be impossible to calculate with the information given. This speed also assumes no velocity lost in breaking the window, an obviously false assumption.*

Thanks very much to Craig for taking the time to do this.

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