question from my Physics 1 class (qualitative explanation?)

[Prodigy54321]

I'm learning about projectile motion /images/graemlins/tongue.gif...the question is..

What is the angle of projection from the ground for which the range is a maximum? Can you give a qualitative explanation?

45% obviously..then I wrote out an explanation about how the range is a maximum when sin2(theta) is at it's maximum of 1...2(theta) = 90 degrees....(theta) = 45 degrees

is this considered a "qualitative explanation?"...I think it's more of a quantitative explanation..not sure what my professor is looking for though.

sorry about the stupid question /images/graemlins/smile.gif

[bunny]

Maybe he meant some kind of hand-wavy argument? Like if you start by firing straight up, it doesnt go anywhere. If you reduce the angle to 89,88,etc then it goes further and further. However at horizontal it also doesnt go anywhere, because of gravity. It's kinda linear, so the midpoint is likely to be "optimum".

Having said that, your answer is much better (specially for a physics class) - I really dont see the point in a "qualitative" explanation.

[surftheiop]

I guess you could BS something like
"At 45 degrees half of the energy will go towards keeping the object in the air and half will go towards moving it forward .... and this is like a square ..... and squares maximize things ......"
And hope everyone is impressed (it worked in my highschool physics class)

[AWoodside]

I think it's ridiculous to be asking for a qualitative answer to this question (assuming that's all you're asked for). Just show why it's the case and then write some BS. Given that you have the answer correct and can show it mathematically it would be pretty inane to mark you down for your qualitative description.

[m_the0ry]

At 90 degrees we have maximum airtime. At 0 degrees we have minimum airtime. Since projectile distance is distal velocity times airtime, we maximize the distance at the midpoint between the two just like a thevenin equivalent has max power output when the resistors are balanced.

[Uethym]

Assume that the projectile is launched with velocity v at an angle theta to the ground. The horizontal velocity is then v cos(theta) and the vertical velocity is v sin(theta).

1. Find the amount of time before the projectile hits the ground. Use x = x(0) + v(0)t + (1/2)at^2, where x(0) = 0 (you start on the ground), x = 0 (you finish on the ground), v(0) = vertical velocity, and a = -g (gravitational acceleration). Solve for t.

2. Find how far the projectile travels horizontally in that much time. Use x=vt, where v is horizontal velocity and t is the time you found in part 1.

3. Find the theta that maximizes the distance you found in step 2.

(Edit: I just realized I misread "qualitative" as "quantitative". What a donk. The prof, I mean. What good is a qualitative explanation? Prove it or don't, I say.)

[BruceZ]

The horizontal distance it travels is directly related to 2 things: a) how long it takes to fall to the ground (air time), and b) the velocity in the horizontal direction. The air time is maximized when we shoot it straight up, but then the horizontal velocity is zero. The horizontal velocity is maximized when we shoot it straight forward with no inclination, but then the air time is zero since it hits the ground immediately. Since hang time varies from 0 to 90 degrees in the same way that horizontal velocity varies in the opposite direction from 90 to 0 degrees, the distance is maximized when both factors contribute equally, which occurs at 45 degrees.

[Prodigy54321]

thanks for the help guys