The question is as follows:
Ignoring drag, a golf ball launched with neutral spin can be expected to follow a curve given by:
r(t) = (Vcos$\theta$t, Vsin$\theta$t - g$t^2$/2)
where V: launch speed, theta: launch angle, g: acc due to gravity, t: time
A simple model to incorporate lift force on the ball mass due to backspin is proposed as:
$F_{lift}$ = m$a_1$
....which switches off when t exceeds half the flight time (TOF) for the spin shot to simulate the ball slowing down. a1 is assumed constant. How much will the length of the golf shot be changed when backspin is considered?
My answer was : $\frac{2}{1 - \frac{a_1}{2g}}$ however I'm not sure if it's correct.
My procedure:
Derived a composite function for the vertical y displacement of the ball w/spin for the first half (see bottom) then w/o spin for the second half.
Into each equation, subbed in $\frac{TOF_{spin}}{2}$, summed the displacements and set = 0. (Overall y displacement is 0). Solved for $TOF_{spin}$.
Solved for $TOF_{nospin}$ then found ratio of TOFs and thus ratio of ranges as range=u.TOF
Any input would be greatly appreciated!
- ð $r_{y-spin}$(t) = Vsin$\theta$t + $\frac{1}{2}$($a_1$-g)$t^2$
Calculations:
However I think that I cannot sum the formulae as I have done here:
Your approach is generally correct, but there's one problem: your flight path is discontinuous. For the $y$-coordinate of the ball, you wrote \begin{equation} y(t) = \left\{ \begin{array}{rcl} V\sin \theta\,t + \frac{1}{2}(a_1-g)t^2 & \text{if} & 0 < t \leq \frac{1}{2}T, \\ V\sin \theta\,t -\frac{1}{2}g t^2 & \text{if} & \frac{1}{2}T < t \leq T\end{array}\right. . \end{equation} However, at $t=T/2$, the ball suddenly drops from height $V \sin \theta T + \frac{1}{2}(a_1 - g) T^2$ to height $v \sin \theta T - \frac{1}{2} g T^2$. In other words, the second half of your flight path is not correct.
To remedy this, consider this question: what would the flight path be of a golf ball, starting at the coordinate $(x_0,y_0)$, launched with initial speed $W$ at an angle $\phi$? Note that your 'neutral' flight path belongs to a golf ball launched from the origin (i.e. $(x_0,y_0) = (0,0)$), at an angle $\theta$ with initial speed $V$.