2019-01-28 ~ 2019-01-29
### Field Goal Velocity and Angle

*MIT Technology Review* recently asked for the minimum initial
velocity and associated vertical kick angle for a field goal attempt
of horizontal distance :(x): to strike the crossbar at height :(y):.

Let :(v): and :(\theta): be the initial velocity and angle, respectively. Then the horizontal and vertical travels of the ball at time :(t): after the kick are :[ \eqalign { x &: t\mapsto vt\cos\theta, \cr y &: t\mapsto vt\sin\theta-\frac12 gt^2, \cr } ]: respectively, where :(g): is the acceleration due to gravity. Solving the first equation for :(t):, substituting into the second, solving for :(v):, and zeroing the derivative with respect to :(\theta): produces the minimum velocity :(v^*): by way of the associated angle :(\theta^*):: :[ \eqalign { v^* &= \sqrt{g\left(y+\sqrt{x^2+y^2}\right)}, \cr \theta^* &= \frac{\pi-\arctan(x/y)}2. \cr } ]: Here's what this looks like for typical NFL conditions:

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Let :(v): and :(\theta): be the initial velocity and angle, respectively. Then the horizontal and vertical travels of the ball at time :(t): after the kick are :[ \eqalign { x &: t\mapsto vt\cos\theta, \cr y &: t\mapsto vt\sin\theta-\frac12 gt^2, \cr } ]: respectively, where :(g): is the acceleration due to gravity. Solving the first equation for :(t):, substituting into the second, solving for :(v):, and zeroing the derivative with respect to :(\theta): produces the minimum velocity :(v^*): by way of the associated angle :(\theta^*):: :[ \eqalign { v^* &= \sqrt{g\left(y+\sqrt{x^2+y^2}\right)}, \cr \theta^* &= \frac{\pi-\arctan(x/y)}2. \cr } ]: Here's what this looks like for typical NFL conditions:

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