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#

1 Answer
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.

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#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)#