# How to find x(t) and y(t)?

## Consider the trajectory of a golf ball which will be hit with a club, where the initial speed of the ball is ${v}_{o}$ and the angle at which the golf ball leaves the golf club is $\alpha$. Assume that the horizontal acceleration is ${a}_{x} = - k {v}_{x}^{2}$ (where ${v}_{x}$ is the horizontal speed) and vertical acceleration is only due to gravity $g$. Find expressions for x and y positions of a ball as a function of time $t$, in terms of the initial conditions for the ball ${v}_{o}$, $\alpha$, and the aerodynamic drag coefficient $k$. Answer: $x \left(t\right) = \frac{1}{k} \ln \left(k t {v}_{o} \cos \alpha + 1\right)$, $y \left(t\right) = - \frac{1}{2} g {t}^{2} + {v}_{o} t \sin \alpha$

May 21, 2016

As below.

#### Explanation:

Let us look at the following figure to see the initial conditions.
It does not depict the actual trajectory followed by the golf ball.

$x$ component of the initial velocity ${v}_{\circ} = {v}_{\circ} \cos \alpha$
$y$ component of the initial velocity ${v}_{\circ} = {v}_{\circ} \sin \alpha$

Since both $x \mathmr{and} y$ components are orthogonal to each other, hence both can be treated independently.

1. Displacement along $y$ axis.
We may use the kinematic equation
$s = u t + \frac{1}{2} a {t}^{2}$, ......(1)
where $s$ is the distance moved in time $t$, $u$ is initial velocity and $a$ is the acceleration. Inserting given values we obtain
$y \left(t\right) = \left({v}_{\circ} \sin \alpha\right) t - \frac{1}{2} g {t}^{2}$
$-$ sign in front of acceleration due to gravity $g$ shows that it is in $- y$ direction or opposite to the direction of displacement in the $y$ direction. Rearranging

$y \left(t\right) = - \frac{1}{2} g {t}^{2} + {v}_{\circ} t \sin \alpha$

1. Displacement along $x$ axis

$x \left(t\right) = {\int}_{0}^{t} {v}_{x} . \mathrm{dt}$ ....(2)

Now we are to find out ${v}_{x}$ as a function of $t$

We know that ${a}_{x} = - k {v}_{x}^{2}$

So $\frac{{\mathrm{dv}}_{x}}{\mathrm{dt}} = - k {v}_{x}^{2}$
or $\frac{{\mathrm{dv}}_{x}}{{v}_{x}^{2}} = - k \mathrm{dt}$
Integrating both sides

$\int \frac{{\mathrm{dv}}_{x}}{v} _ {x}^{2} = - k \int \mathrm{dt}$

$\implies {v}_{x}^{- 2 + 1} / \left(- 2 + 1\right) = - k t + c$, where $c$ is integration constant.

$\implies {v}_{x}^{- 1} / \left(- 1\right) = - k t + c$ ....(3)

Imposing boundary condition: at $t = 0 , {v}_{x} = {v}_{\circ} \cos \alpha$
From (3) we have
$- \frac{1}{{v}_{0} \cos \alpha} = c$

Inserting this value of $c$ in equation (3) we get

$- \frac{1}{v} _ x = - k t - \frac{1}{{v}_{\circ} \cos \alpha}$

$\implies {v}_{x} = \frac{{v}_{\circ} \cos \alpha}{k t {v}_{\circ} \cos \alpha + 1}$

Now inserting this value of ${v}_{x}$ in equation (2) we get

$x \left(t\right) = {\int}_{0}^{t} \frac{{v}_{0} \cos \alpha}{k t {v}_{\circ} \cos \alpha + 1} \mathrm{dt}$ ....(4)

Now let $z = k t {v}_{\circ} \cos \alpha + 1$ .....(5)

when $t = 0$ then $z = 1$

Differentiating (5) w.r.t to respective variables we get

$\mathrm{dz} = k {v}_{\circ} \cos \alpha \mathrm{dt}$

$\implies {v}_{\circ} \cos \alpha \mathrm{dt} = \frac{1}{k} \mathrm{dz}$

Now equation(4) becomes

$x \left(t\right) = {\int}_{1}^{k t {v}_{\circ} \cos \alpha + 1} \frac{1}{k} \frac{\mathrm{dz}}{z}$

$= \frac{1}{k} \ln z |$, within limits as above. Constant of integration has been omitted due to definite integral and as $\ln 1 = 0$

$\implies x \left(t\right) = \frac{1}{k} \ln \left(k t {v}_{\circ} \cos \alpha + 1\right)$