Posted by **TK** on Monday, September 1, 2008 at 4:33am.

This is going to be hard to explain, but I did an experiment on projectile motion and we had to lauch a marble from 0 degrees to 90 degrees in a 5 degree interval using a projectile launcher with the same velocity each time.

We had to investigate the angle which gave the greatest range, and I graphed the results. The graph looked like the letter M, where range was on the y-axis, and the degree of launch was on the x-axis. The graph increased until 30 degrees, then there was a slight drop, then it peaked again at 50 degrees.

THe question asked

** Use your graph to determine the relationship between the range of a projectile and pairs of launch angles, eg 15 degrees and 75 degrees **
I would think the answer would be a quartic relationship by looking at the graph of some sort, but then again we haven't learnt quartic yet. I think it could be also a quadratic realtionship because projectiles move in a paranolic matter, and I think my teacher mentioned it once before.

- Physics - Projectile Motion -
**Damon**, Monday, September 1, 2008 at 5:48am
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

so

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

for range

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

SO

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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