Posted by TK on Monday, September 1, 2008 at 4:33am.
Ignoring the friction of the air, the parabola is what it should do.
The 50 degrees is not bad. I do not know how your setup worked but suspect that anywhere from 30 to 60 degrees might look good with the inevitable uncertainties in launch speed.
If you did it with math instead of experiment you would come up with 45 degrees.
u = Uo forever in the horizontal direction
v = Vo - 9.8 t in the vertical direction with initial speed Vo up and time t and acceleration of gravity = 9.8 m/s
tan A = Vo/Uo tangent of launch angle to horizontal
s^2 = Uo^2+Vo^2 = speed at launch
x = Uo t distance down range
y = Vo t - 4.9 t^2
t = x/Uo
y = Vo x/Uo - 4.9 x^2/Uo^2
NOTICE that is a parabola
the projectile comes back to ground when y = 0 so
4.9 xmax^2/Uo^2 = Vo x/Uo
4.9 xmax = Vo Uo
xmax = Vo Uo / 4.9
remember tan A = Vo/Uo so Vo = Uo tan A
so xmax = Uo^2 tan A/4.9
for a given s, what is the maximum range?
Uo^2 = s^2 cos^2 A
so xmax = s^2 cos^2 A (sin A/cos A)
= s^2 sin A cos A
our maximum range is where sin A cos A is maximum
Now you can graph that or use calculus
graphing sin A cos A is hard on the computer so I will use calculus.
f(A) = sin A cos A
dF/dA = sin A (-sin A) + cos A (cos A)
for max or min df/dA = 0 so
sin^2 A = cos^2 A
that is true when A = 45 degrees
So you should have found maximum range when A is 45 degrees which is why I said 50 degrees was pretty good.
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