
#1
Jan1013, 10:29 AM

P: 197

Can anyone explain to me why (when air resistance is ignored), that 45° angle is the ideal angle for the longest range? Is it because if it is over 45° then gravity will be acting on it longer therefore will be dragged down sooner? Also as "u cos >45" produces a lower exit velocity(horizontally)? But what if it is below 45? Then the exit velocity is greater than at 45, but then does the greatest height have something to do with it does travel as far as when the angle is closer to 45.
Also when air resistance is taken into account, there is a force acting in the opposite direction to the velocity, so obviously the range will not be as great, however, how come there is a bigger difference between the ideal and actual ranges at angles above 45 than below? Is it due to f=ma and the fact that the projectile is travelling more in the vertical direction and that the exit velocity would not be as great(horizontally) so it does not take as much force to slow it down than angles lower than 45? 



#2
Jan1013, 01:17 PM

P: 742

The horizontal velocity controls how much progress the projectile can make per unit time. The higher the velocity the farther it can get in a fixed time.
The vertical velocity controls how much time it will take before the projectile hits the ground. The higher the velocity the more time before it lands. The goal is to maximize total distance. So you want to maximize the product of horizontal velocity and time. Clearly a purely vertical trajectory is suboptimal. You maximize time, but the resulting distance is zero. Clearly a purely horizontal trajectory is suboptimal. You maximize horizontal velocity, but the resulting distance is, again, zero. The horizontal velocity scales as the sine of the elevation above the horizontal The vertical velocity scales as the cosine of the elevation above the horizontal. You can apply the halfangle formula and derive that the maximum must be at 45 degrees because sin(theta)cos(theta)/2 = sin(2 theta) and sin(2 theta) is maximized when 2 theta = 90 degrees so theta = 45 degrees. Or you can notice the symmetry and realize that if there is a single maximum then it must occur at 45 degrees. Or you can look at the first derivitive of sin(theta)cos(theta). By the product rule, that's ( sin(theta) * d cos(theta) + cos(theta) * d sin(theta) ) / d theta And that's cos^2(theta)  sin^2(theta). The maximum is obtained when this result is zero. So cos^2(theta) = sin^2(theta). In the first quadrant, that means that cos(theta) = sin(theta). By inspection that occurs at theta = 45 degrees. 



#3
Jan1013, 02:58 PM

P: 3,536

To put it more precisely:
1) The horizontal distance is proportional to the constant horizontal velocity and to the flight time. 2) Since the vertical acceleration is constant, the flight time is proportional to the initial vertical velocity. 1 + 2) The horizontal distance is proportional to the product of the horizontal and vertical velocity. Now think of a rectangle with the sides Vx and Vy. Given a fixed diagonal V, the area Vx * Vy is maximized in the case of a square: Vx = Vy or 45° diagonal inclination. 



#4
Jan1113, 05:24 PM

P: 197

Projectile motion rangeAlso, I assume its not going as far because the vertical exit velocity is greater than the horizontal, am I correct in assuming that? Thank you! 



#5
Jan1213, 03:28 AM

P: 312

Well the maths saids so..




#6
Jan1213, 05:55 AM

P: 197

What about when air resistance is taken into account, is there a reasons for a bigger difference in actual and ideal ranges at angles > 45° than < 45°? This one has puzzled me. At the basic level, I think it has something to do with f=ma and the fact its travelling more in the vertical direction but cant seem to figure out why? Any help is greatly appreciated. 



#7
Jan1313, 08:18 AM

P: 197

Can anyone please help with the air resistance thing?
Have a look at the image I can see there that if the angle were greater than 45, than the y component of the air resistance would be greater meaning more force pushing it down. Therefore, I can get why there is a big difference between actual and ideal at angles greater than 45(sort of). But what about when the projectile is coming back down? The air resistance would be pushing it back (like the y component of the Fair is pushing it back into the air), would that negate the effects described in the above paragraph or not? 



#8
Jan1313, 04:14 PM

P: 3,536





#9
Jan1413, 11:49 PM

P: 197

Can anyone tell me if I am correct in making the assumption about the air resistance as per the image above, when the angle is greater than 45°, the y component of the air resistance would be greater than its x, therefore acting downwards it now has gravity and most of the air resistance, causing it to decelerate towards the ground faster than if it was shot at angles below 45°? I can see it both ways, as if the angle was below 45° the x component of the air resistance is greater than its y, but then it has gravity acting down, and the air resistance simply pushing against it, which would still slow it down. But I am not sure if it would slow it down just the same as above or not as fast? Thanks. 



#10
Jan1513, 04:19 AM

Sci Advisor
PF Gold
P: 11,341

As A.T. said, why bother talking and arm waving when the Maths says it all so much more precisely? If every bit of Maths needed to be restated in conversational terms, we'd never have any advanced Physics. Maths is an appropriate language for this sort of problem so let's use it.




#11
Jan1613, 05:22 AM

P: 3,536

All the other optimal cases are somwhere between the two extremes: 45° for drag = 0 ~0° for drag > [itex]\infty[/itex] 


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