For a given initial projectile speed, you observe that the projectile has a certain range R at a launch angle A = 30 degrees. For what other launch angle will the projectile have the same range (assuming the same initial projectile speed)? Without the use of mathematical formula give a conceptual argument as to why this angle gives the same range (think about the tradeoffs between time-of-flight and horizontal component of velocity in your answer).

2 answers

shoot straight at the target with a little up angle for drop or shoot almost straight up so it gets the target on the way down. There are two solutions for the equations that yield time for given height of target.
( Fire the ball or lob it :)

Vi = Vo sin T
u = Vo cos T

v = Vi - 9.8 t
v at top = 0
t = Vi/9.8 = (Vo/9.8)sin T
2 t = time in air (Vo/4.9) sin T

range = u (2 T)
= Vo cos T (Vo/4.9) sin T
= (Vo^2/4.9) cos T sin T

if T = 30
sin T = .5 and cos T = .866
but if T = 60
sin T = .866 and cos T = .5
same range
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