Monday

December 22, 2014

December 22, 2014

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

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

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 5:48amIgnoring 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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