General Projectile Question about solving 2-d PHysics problem?

Question

Am I correct in thinking that angles will reach their maximum range at 45 degrees....therefore, when applying this concept to projectile motion problems, one should...

...take the original angle provided-for example 60 degrees-double it, and find the difference between 60 doubled (120) and 45 doubled (90); 120-90=30; therefore, the angle from which an alternative projectile can be shot to arrive at the same horizontal point is 30 degrees??

...This is how I understood someone's explanation to another physics problem that I asked (thanks to the person who took the time to answer!)...

...Anyway, I wanted to make sure I understood this concept correctly...does this apply to all angles from which projectiles are shot??

Thank you!!!

OH-I just wanted to add that I calculated the 30 degrees by subtracting the 30 from 60 degrees...

Answer 1

I think that what you have said is correct for a very specific problem, but if you think of it in this narrow way, you will be woefully restricting your way of thinking about trajectories.

Here's how it works:
Initial velocity is Vo at angle A from the horizontal, so initially
v-x(0) = Vo cos(A)
v-y(0) = Vo sin(A)

Over time, v-y will go to 0 and go negative; it will have the same value as before (but negative) when it hits the same height. That happens when:
v-y(t) = Vo sin(A) - gt is equal to = - Vo sin(A)

or gt = 2Vo sin(A)
t = 2Vo sin(A)/g

Along the x axis, the distance traveled has been:
x(t) = v-x * t (because v-x has not been changing with time)
= Vo cos(A) t
= Vo cos(A) 2 Vo sin(A) /g
=2 (Vo^2) cos(A) sin(A) / g
= sin(2A) Vo^2)/g

So it's true that the range is greatest when Y = 45 degrees.
And it's true that you'll get the same range when you tilt it up from 45 degrees or when you tilt it down from 45 degrees by the same amount (which is equivalent to what you were saying).

But what if they change the problem? Suppose you are shooting the cannon from a tower that is B meters higher than the field where the ball is to land? In that case, the final range won't work out neatly in terms of sin(2A), so this "rule of thumb" will lead you astray.

Answer 2

You are correct, sir! If two angles are equally spaced from 45°, the projectile will land at exactly the same distance. 60° and 30° are two such angles, and the calculation you describe is a way to find out what angle is the "mirror-image" of the other angle across the 45° line.

This is all of course true only in the absence of air-resistance. It took them many centuries to figure out the effect of air-resistance, which is a force proportional to the speed of the projectile but since speed was also a function of gravity they were stumped, until they could find a way to use equations with both quantities and their rates of change at the same time, i.e. the calculus of differential equations.

Answer 3

Hi,

Projectiles shot at a 45° angle will have the longest range. Angles that are ±2° from 45° will have the same range, so 43° and 47° have the same range. In the same way 40° and 50° have the same range(±5°), and 26° and 64° have the same range because they are ±19° from 45°.

I hope that helps!! :-)

Answer 4

The general equation launched at an angle z from the origin is y = -gx^2/(2Vo^2 cos^2 z ) + x tan z. This is obviously a parabola. Vo = initial velocity, g is the gravitational constant, and z is the inclination angle.

Parametrically, the equation using time t as a parameter:
x = (Vocos z)t and y = -1/2 gt^2 + (Vosin z)t, which give the x and y coordinates at any time t seconds after launch.

This should be enough to let you investigate farther.

Answer 5

calculus is more difficult than another math...if u havent taken physics ever, i recommend u to not take each in combination, except u wish to diminish ur g.p.a., i attempted taking each in combination this semster n its quite tough, if u wish to go take them each seperatly so u might cognizance then the opposite... well success!!!