## Problem:An airplane fying at a height h above the ground is suddenly hit by a shell fired from the ground. It is known that the shell was fired at mozzle speed v0 from the ground but the exact position and angle of elevation of the gun are unknown. Deduce that the gun is situated within a circle whose center lies directly below the plane and whose radius is |

> #

2 1/2 v0 (v0 - 2 g h) ------------------- g

## Solution:We need to find the maximum value of R (for radius) that still hits the plane flying at altitude h. Where, |

> R := v0*cos(a)*t;

R := v0 cos(a) t> h := v0*sin(a)*t - g*t^2/2;

2 h := v0 sin(a) t - 1/2 g t

From here and the equation for R we obtain, |

> E1 := ('h'+g*t^2/2)^2 = v0^2*t^2-'R'^2;

2 2 2 2 2 E1 := (h + 1/2 g t ) = v0 t - R

Hence, |

> 'R'^2 = v0^2*t^2-('h'+g*t^2/2)^2;

2 2 2 2 2 R = v0 t - (h + 1/2 g t )

We now find the value of t that maximizes this expression. Since t >0 it is convenient to, |

> h := 'h': F := z -> v0^2*z-(h+g*z/2)^2;

2 2 F := z -> v0 z - (h + 1/2 g z)

The max of F = R^2 is at z=t^2 obtained when, |

> diff(F(z),z) = 0;

2 v0 - (h + 1/2 g z) g = 0

i.e. when, |

> z := solve(diff(F(z),z),z);

2 -v0 + g h z := -2 ---------- 2 g

substituting in the expression for R^2 we obtain, |

> R := 'R': R^2 = F(z);

2 2 / 2 \2 2 v0 (-v0 + g h) | -v0 + g h| R = -2 ---------------- - |h - ----------| 2 \ g / g

this simplifyies to: |

> R^2 = factor(expand(F(z)));

2 2 2 v0 (-v0 + 2 g h) R = - ------------------ 2 g

and taking sqrt on both sides we obtain the desired result: |

> R = v0*sqrt(v0^2-2*g*h)/g;

2 1/2 v0 (v0 - 2 g h) R = ------------------- g

Link to the commands in this file

Carlos Rodriguez <carlos@math.albany.edu> Last modified: Wed Oct 10 11:13:19 EDT 2001