A physic question - very interesting but hard?

Question

The question is in the image
http://img300.imageshack.us/img300/5259/question01em4.jpg

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Please explain in detail, I very much appreciate your help

Answer 1

Consider each component of the velocity of the arrow separately.

The horizontal speed is
Vb*cos(th)+Vh

The range of the shot will be
d=t*(Vb*cos(th)+Vh)
now we must find t which is the flight time.

Under the ideal conditions stated
Vb*sin(th)*t-g*t^2/2=y(t)
When the arrow leaves the bow and when the arrow lands at d y(t)=0

Therefore
t=Vb*sin(th)*2/g

Plug in above
d=(Vb*sin(th)*2/g)*
(Vb*cos(th)+Vh)

Part b requires differentiation using the product rule
d d/d th
=vb*cos(th)*2/g*(vb*cos(th)+vh)-
vb^2*sin^2(th)*2/g
set equal to 0 and solve for th
(note that without the horse the solution is cos^2-sin^2=0)
0=cos(th)*(vb*cos(th)+vh)-
vb*sin^2(th)
since
1=cos^2+sin^2
sin^2=1-cos^2

0=vb*cos^2(th)+cos(th)*vh-
vb+vb*cos^2(th)
0=2*vb*cos^2(th)+cos(th)*vh-vb

so cos(th) can be solved as a quadratic
set up a variable v=vh/(2*vb)
0=cos^(th)+v*cos(th)-.5
cos(th)=(-v+/-sqrt(v^2+2))/2

I looked at some examples I thought were reflective of reality.
A fast horse runs 30 to 40 mph. A slow arrow flies at 100 mph
at 30 mph for the horse, that gives a v=.15 and th = 50.5
at 40 and 100 v=.2, th=52.1. Note how as the horse goes faster the angle increases (btw, since it is quadratic there are two roots, the other root gives the angle for maximum distance if the archer shot backwards away from the direction of the horse. The angles are more shallow since the speed of the horse reduces the horizontal component)

Intuitively we know that th should be greater than 45 for a forward shot since the horse adds horizontal component of velocity so the flight time will be longer, therefore you want the horizontal and vertical components to be equal

ie
vh+cos(th)*vb=sin(th)*vb
This is only true for angles greater than 45 degrees


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