Projectile angle for maxium distance
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because wiith the 45 degree angle we are going to have an optimal blend between horizontal and vertical velocities
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Don't know what combining horizontal and vertical velocities have to do with this, but...
Assuming distance in the x direction, then: d=(cos Vx)(t). Clearly from this equation there is a direct coorelation between the magnitude of vector Vx and d, therefore the higher values for cos should be the answer. If we think about this conceptually, throwing a pitch side arm almost parallel to the ground (for max value of cos=1 at angle of 0 degrees) would yield a further x distance than throwing it up in the air at a lesser angle--assuming equal velocities, of course.
Assuming distance in the x direction, then: d=(cos Vx)(t). Clearly from this equation there is a direct coorelation between the magnitude of vector Vx and d, therefore the higher values for cos should be the answer. If we think about this conceptually, throwing a pitch side arm almost parallel to the ground (for max value of cos=1 at angle of 0 degrees) would yield a further x distance than throwing it up in the air at a lesser angle--assuming equal velocities, of course.
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According to your explanation then throwing it at a 30 degree angle would be better than throwing it at a 45 degree angle, which is just untrue. We want to look at both horizontal and vertical velocities because the vertical velocity will dictate time of flight.
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I agree with IntelInside... a 45 degree angle is the optimal blend. Here's an example (slightly modified) taken from EK 1001
A projectile is launched at an angle of 30 degrees to the horizontal. It's initial velocity is 20m/s. How far does it travel?
To solve this you have to first find total flight time. Using v=vo+at where v=0m/s at the top of the flight path; vo=vo(sin30); and a=-10m/s
We see that t=1 sec
That's only for half the trip though so total flight time is t=2 sec
Now to find total distance traveled we can use the equation d=vot+1/2at^2
Where a=0 because there's no acceleration in the x-direction; t=2 sec; and vo=vo(cos30)
We see that d=36m
Now, do the same problem with a 45 degree angle.
We see that t=3.4 sec and d=57.8m
Well, what about a 60 degree angle?
We see that t=3.6 sec and d=36m
A 45 degree angle is clearly the magic angle that maximizes travel time and distance.
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Vf=Vo+a*t
If dy=0 then Vyf=-Vy0.
Vf= -(Vf)+a*t
2 Vf = a*t
t = 2 VYf / a
VYf = V*sin(x)
t = 2 * V * sin (x) /a
Dx = V * cos (x) * t
= V * cos (x) * 2 * V * sin (x) /a
= V² / a * 2 * cos (x) * sin (x)
Clearly you want to maximize the product of cos(x) and sin (x) to get the maximum displacement.
Or if you are familiar with trigonometric identities then you will notice sin (2x) = 2 sin (x) cos (x)
so that the prior equation reduces to V² / a * sin (2x)
and since sin(x) reaches a max at 90° the equation V² / a * sin (2x) reaches a maximum at 45°.
Of course this all assumes zero vertical displacement.
If you use 30 degree, then projectile will land too soon comparing to 45 deg., which aka. less distance. You can also confirm the answer by setup a right triangle, and figure out what angles make the longest hypotenuse.
Here's the explanation in a nutshell:
Distance traveled = velocity x time
A low angle(e.g. cos(30)) would give you a high velocity, but it would give you a small value for time since the time a projectile spends in the air is related to the sine of the angle(e.g. sin(30))
A high angle would give you a high value for time(e.g. sin(60)), but it would give you a small velocity since the velocity of a projectile is related to the cosine of the angle(e.g. cos(60)).
The magical mix, where velocity x time is maximum, occurs at 45 degrees, when cosine = sine.
This is a similar train of thought as why a square will have a larger area than a rectangle with an identical perimeter, or why 5x5 > 4x6 > 3x7 > 2x8....etc.
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increasing the angle will increase the time and vertical displacement of the projectile motion as these components depend on sine.
horizontal displacement depends on both sine and cosine. thus, at 45 degrees you get the optimal angle for maximum horizontal displacement.
for visual learners: http://www.physicsclassroom.com/mmedia/vectors/mr.cfm
horizontal displacement depends on both sine and cosine. thus, at 45 degrees you get the optimal angle for maximum horizontal displacement.
for visual learners: http://www.physicsclassroom.com/mmedia/vectors/mr.cfm
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i hate physics. but as everyone said ahead of me just simplified, keep in mind that velocities of objects in the x and y directions independently of each other and 45 degrees gives you the best ratio of velocity in y direction (important for keeping the ball, bullet, cannonball, kitten with cancer etc in flight for a longer length of time) to velocity in the X direction resulting in the furthest distance traveled.
I agree, it is about a optimal combination of vertical and horizontal velocities.
The higher it goes, the longer it takes for it to fall to the ground, but the more velocity you give it in the vertical direction, the less velocity it has to go horizontally.
Horizontal velocity stays constant through the trajectory (ignoring air resistance), but the acceleration downwards (gravity) pulls the object down, and once it hits the ground, it obviously aint going anywhere.
The higher it goes, the longer it takes for it to fall to the ground, but the more velocity you give it in the vertical direction, the less velocity it has to go horizontally.
Horizontal velocity stays constant through the trajectory (ignoring air resistance), but the acceleration downwards (gravity) pulls the object down, and once it hits the ground, it obviously aint going anywhere.
