Question

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=-kv_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(t)=1/kln(ktv_ocosalpha+1)`, `y(t)=-1/2g t^2+v_otsinalpha`

Answer 1

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.

![wikimedia.org]()
`x` component of the initial velocity `v_@=v_@cos alpha`
`y` component of the initial velocity `v_@=v_@sin alpha`

Since both `x 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=ut+1/2at^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(t)=(v_@sin alpha)t-1/2g 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(t)=-1/2g t^2+v_@ tsin alpha`

1. Displacement along `x` axis

`x(t)=int_0^tv_x.dt` ....(2)

Now we are to find out `v_x` as a function of `t`

We know that `a_x=-kv_x^2`

So `(dv_x)/(dt)=-kv_x^2`
or `(dv_x)/(v_x^2)=-kdt`
Integrating both sides

`int(dv_x)/v_x^2=-kintdt`

`=>v_x^(-2+1)/(-2+1)=-kt+c`, where `c` is integration constant.

`=>v_x^(-1)/(-1)=-kt+c` ....(3)

Imposing boundary condition: at `t= 0, v_x=v_@cosalpha`
From (3) we have
`-1/(v_0cosalpha)=c`

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

`-1/v_x=-kt-1/(v_@cosalpha)`

`=>v_x=(v_@cosalpha)/(ktv_@cosalpha+1)`

Now inserting this value of `v_x` in equation (2) we get

`x(t)=int_0^t(v_0cosalpha)/(ktv_@cosalpha+1)dt` ....(4)

Now let `z=ktv_@cosalpha+1` .....(5)

when `t= 0` then `z=1`

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

`dz=kv_@cosalphadt`

`=>v_@cosalphadt=1/kdz`

Now equation(4) becomes

`x(t)=int_1^(ktv_@cosalpha+1) 1/k(dz)/(z)`

`= 1/k lnz|`, within limits as above. Constant of integration has been omitted due to definite integral and as `ln1=0`

`=>x(t)=1/kln(ktv_@cosalpha+1)`