You kick a ball over a flat field. Taking into account gravity, but disregarding everything else like wind, friction, bounces, etc., etc., at what angle should you kick it so the ball lands the farthest away from you? And at what angle should you kick it so the ball makes the longest trajectory before landing?
Assume that initial velocity V is what is constant and all you can vary is the angle from horizontal, which I'm calling B (since I can't make a theta)
After the kick, there is no more acceleration except gravity.
x(t)=V cosB t
y(t)=V sinB t  (1/2) g t^2
y'(t)=V sinB  g t which = zero at vertical peak of trajectory
ie when t=(V sinB / g)
so entire trip time is twice that: 2 V sinB / g = t(final)
Proof of part 1:
x(max) = V cosB t(final) = 2 V^2 sinB cosB / g = V^2 sin(2B) / g
which has a max value of V/g iff sin(2B)=1, so B= 45 degrees
Part 2: Trajectory S = integral {t=0 to t(final)} ds
where ds^2 = dx^2 + dy^2
dx= V cosB dt
dy= [V sinB  gt] dt
(ds/dt)^2= V^2 (cosB)^2 + V^2 (sinB)^2 +(gt)^2  2Vg sinB t
+ 2 V^2 sinB cosB  2Vg cosB t
= g^2 t^2  2Vgt (sinB+cosB) + V^2 (1+ sin(2B))
which is a perfect square and factors to
=[gt  V(sinB+cosB)]^2
(easiest way to see this is to show that b^24ac is zero)
So ds/dt = gt  V(sinB+cosB) absolute value since it has to be positive
S =  integral {t=0 to t(final)} [gt  V(sinB+cosB)] dt 
=  (1/2) g t^2  V t (sinB+cosB) evaluated at the t values 
=  2 V^2 (sinB)^2 /g  2 V^2 sinB (sinB+cosB) / g 
S = V^2 sin(2B) / g
So, for part 2, I'm getting the same answer for angle B that I got in part 1:
x(final)=V^2 sin(2B) / g
S(final)=V^2 sin(2B) / g
and each is maximized if sin(2B)=1 which means B=45 degrees
Herein lies my problems:
(1) How can the max x distance and the trajectory be the same?
(they can't)
(2) Charlie is never wrong,
..... so where did I go wrong?

Posted by Larry
on 20050319 23:03:39 